A cylindrical cork of uniform density floats in a liquid of density $\rho_1$. If the cork is depressed slightly and released, it oscillates harmonically with time period $T$. If the same cork floats in another liquid of density $\rho_2$, then the similar oscillation has time period $2T$. The value of $\rho_2/\rho_1$ is :
Concept. Pushing a floating cork down by $y$ adds a buoyant restoring force $\rho Agy$, so $k = \rho Ag$ and $T = 2\pi\sqrt{\dfrac{m}{\rho Ag}} \propto \dfrac{1}{\sqrt\rho}$.
Step 1. Ratio of periods. $$\frac{T_2}{T_1} = \sqrt{\frac{\rho_1}{\rho_2}} = 2$$
Consider a spring-mass simple harmonic oscillator in one dimension. The mass of the particle is $m$ kg and the spring constant is $k\ \text{N m}^{-1}$. At a given instant, the extension of the spring is $x$ meter and the speed of the particle is $v\ \text{m s}^{-1}$. On the $x$ - $v$ plane, if the graph of $v$ as a function of $x$ is a circle, then the correct option is :
Concept. Energy conservation gives the $x$–$v$ curve: $\dfrac12mv^2 + \dfrac12kx^2 = \dfrac12kA^2$.
Step 1. Curve. $x^2 + \dfrac{m}{k}v^2 = A^2$, an ellipse in general.
Step 2. Circle condition. The coefficients must be equal: $\dfrac{m}{k} = 1$, i.e. $k = m$ (numerically, in SI units; $\omega = 1$ rad/s).
Answer. $k = m$ (Option B).
Solution by Sreeraj P, M.Sc Physics
Q 11-13-003NEETNEET 2025Top questionHard
In an oscillating spring mass system, a spring is connected to a box filled with sand. As the box oscillates, sand leaks slowly out of the box vertically so that the average frequency $\omega(t)$ and average amplitude $A(t)$ of the system change with time $t$. Which one of the following options schematically depicts these changes correctly?
Step 2. Amplitude. The sand leaves with the box's own velocity and exerts no horizontal force; with $\omega A^2$ constant, $A$ decreases as $\omega$ increases. (Each grain carries away its own kinetic energy, so the box's energy $\tfrac12kA^2$ falls.)
Two identical point masses P and Q, suspended from two separate massless springs of spring constants $k_1$ and $k_2$, respectively, oscillate vertically. If their maximum speeds are the same, the ratio ($A_Q/A_P$) of the amplitude $A_Q$ of mass $Q$ to the amplitude $A_P$ of mass $P$ is:
If $x = 5\sin\left(\pi t + \dfrac{\pi}{3}\right)$ m represents the motion of a particle executing simple harmonic motion, the amplitude and time period of motion, respectively, are :
If the mass of the bob in a simple pendulum is increased to thrice its original mass and its length is made half its original length, then the new time period of oscillation is $\dfrac{x}{2}$ times its original time period. Then the value of $x$ is:
Concept. For a spring $kl$ is constant, so each half has $2k$; $T \propto \dfrac{1}{\sqrt k}$.
Step 1. New period. $T' = \dfrac{T}{\sqrt2}$.
Answer. $\dfrac{T}{\sqrt2}$ (Option B).
Solution by Sreeraj P, M.Sc Physics
Q 11-13-021NEETJEE MainMedium
A block hangs from a light spring, which stretches it by $4$ cm at equilibrium. If the block is pulled down slightly and released, its period of oscillation is ($g = \pi^2\ \text{m/s}^2$)
A $1$ kg block on a spring of force constant $100$ N/m oscillates with frequency $f$. What mass must be added so that the frequency becomes $\dfrac{f}{2}$?
Concept. If one side rises by $y$, the level difference $2y$ gives a restoring force $2\rho gAy$ acting on the whole column mass $\rho AL$.
Step 1. Equation of motion. $$\rho AL\ddot y = -2\rho gAy \;\Rightarrow\; \omega^2 = \frac{2g}{L}$$
Step 2. Period. $T = 2\pi\sqrt{\dfrac{L}{2g}}$.
Answer. $2\pi\sqrt{\dfrac{L}{2g}}$ (Option B).
Solution by Sreeraj P, M.Sc Physics
Q 11-13-029JEE MainHard
A uniform rod of length $L$ is pivoted at one end and oscillates in a vertical plane as a compound pendulum. The length of a simple pendulum with the same period is
Concept. Compound pendulum: $T = 2\pi\sqrt{\dfrac{I}{mgd}}$; compare with $2\pi\sqrt{\dfrac{l}{g}}$.
Step 1. Rod about its end. $I = \dfrac{mL^2}{3}$, $d = \dfrac{L}{2}$: $$T = 2\pi\sqrt{\frac{mL^2/3}{mgL/2}} = 2\pi\sqrt{\frac{2L}{3g}}$$
Step 2. Equivalent length. $\dfrac{2L}{3}$.
Answer. $\dfrac{2L}{3}$ (Option C).
Solution by Sreeraj P, M.Sc Physics
Q 11-13-030JEE MainHard
A block of mass $M$ on a smooth floor is attached to a spring and oscillates with amplitude $A$. When it passes through the mean position, a lump of putty of mass $m$ is dropped vertically onto it and sticks. The new amplitude is
A tunnel is dug along a diameter of the Earth. A ball dropped into it would perform SHM. Its period is about (radius of Earth $= 6.4 \times 10^6$ m, $g = 10\ \text{m/s}^2$)
A particle executes SHM of amplitude $4$ cm. Its speed at the mean position is $12$ cm/s. Find its speed, in cm/s, when it is $2\sqrt{3}$ cm from the mean position.
Q 11-13-037JEE MainJEE Main 2026 (8 Apr, Shift 2)Easy
The frequency of oscillation of a mass $m$ suspended by a spring is $v_1$. If the length of the spring is cut to half, the same mass oscillates with frequency $v_2$. The value of $v_2/v_1$ is ______.
Q 11-13-038JEE MainJEE Main 2026 (6 Apr, Shift 1)Easy
A particle is executing simple harmonic motion. Its amplitude is $A$ and time period is $5$ sec. The time required by it to move from $x = A$ to $x = \dfrac{A}{\sqrt{2}}$ is ______ sec.
Concept. Use the reference circle. From the extreme ($x=A$, phase $\pi/2$) to $x=A/\sqrt2$ (phase $\pi/4$) the radius turns through $\pi/4$.
Step 1. Fraction of a period. $\dfrac{\pi/4}{2\pi}=\dfrac18$.
Step 2. Time. $t=\dfrac T8=\dfrac58$ s.
Answer. $5/8$ s (Option C).
Solution by Sreeraj P, M.Sc Physics
Q 11-13-039JEE MainJEE Main 2026 (6 Apr, Shift 2)Medium
A spring stretches by $2$ mm when it is loaded with a mass of $200$ g. From equilibrium position the mass is further pulled down by $2$ mm and released. The frequency associated with the system and maxmimum energy in the spring are ______ Hz and ______ J, respectively. (Take $g = 10\ \text{m/s}^2$)
Answer: (A) $\dfrac{5\sqrt{50}}{\pi}$ and $8 \times 10^{-3}$
Concept. The static stretch gives $k=\dfrac{mg}{x_0}$, and $\omega=\sqrt{k/m}=\sqrt{g/x_0}$. The spring's energy is greatest at the lowest point, where its total stretch is $x_0+A$.
Step 1. Spring constant. $k=\dfrac{0.2\times10}{0.002}=1000$ N m$^{-1}$.
Step 2. Frequency. $\omega=\sqrt{\dfrac{1000}{0.2}}=\sqrt{5000}=10\sqrt{50}$; $f=\dfrac{\omega}{2\pi}=\dfrac{5\sqrt{50}}{\pi}$ Hz.
Step 3. Maximum spring energy. Stretch $2+2=4$ mm: $\tfrac12\times1000\times(0.004)^2=8\times10^{-3}$ J.
Answer. Option A. Trap: the SHM energy $\tfrac12kA^2=2\times10^{-3}$ J is not the spring's stored energy.
Solution by Sreeraj P, M.Sc Physics
Q 11-13-040JEE MainJEE Main 2026 (5 Apr, Shift 2)Hard
Match List - I with List - II.
$$\begin{array}{|l|l|l|l|}\hline & \text{List - I} & & \text{List - II} \\ \hline \text{A.} & \sin^2\omega t & \text{I.} & \text{Periodic with time period } T = \frac{\pi}{\omega} \text{ but not simple harmonic motion (SHM)} \\ \hline \text{B.} & \sin^3(2\omega t) & \text{II.} & \text{Periodic with time period } T = \frac{2\pi}{\omega} \text{ but Not SHM} \\ \hline \text{C.} & \sin(\omega t) + \cos(\pi\omega t) & \text{III.} & \text{Periodic with time period } T = \frac{\pi}{\omega} \text{ and SHM} \\ \hline \text{D.} & \cos\omega t + \cos 2\omega t & \text{IV.} & \text{Non-periodic} \\ \hline \end{array}$$
Choose the correct answer from the options given below :
Concept. A motion is SHM only if it can be written as one sinusoid (plus a constant). A sum of sinusoids is periodic only if their frequencies have a rational ratio; the period is then the LCM of the periods.
Step 1. A. $\sin^2\omega t=\tfrac12-\tfrac12\cos2\omega t$: SHM about $\tfrac12$ with $T=\dfrac\pi\omega$. III.
Step 2. B. $\sin^3 2\omega t=\tfrac34\sin2\omega t-\tfrac14\sin6\omega t$: two frequencies, period $\dfrac\pi\omega$, not SHM. I.
Step 3. C. Frequencies $\omega$ and $\pi\omega$: ratio $\pi$ is irrational, so never repeats. IV.
Step 4. D. Frequencies $\omega$ and $2\omega$: period $\dfrac{2\pi}\omega$, not SHM. II.
Answer. Option A.
Solution by Sreeraj P, M.Sc Physics
Q 11-13-041JEE MainJEE Main 2026 (4 Apr, Shift 2)Medium
A uniform disc of radius $R$ and mass $M$ is free to oscillate about the axis $A$ as shown in the figure. For small oscillations the time period is ______.
Concept. Physical pendulum: $T=2\pi\sqrt{\dfrac{I}{Mgd}}$, with $I$ about the axis and $d$ the distance from the axis to the centre of mass.
Step 1. Moment of inertia. About a diameter $I=\tfrac14MR^2$; the axis $A$ is a tangent parallel to a diameter, at distance $R$: $I=\tfrac14MR^2+MR^2=\tfrac54MR^2$.
Q 11-13-042JEE MainJEE Main 2026 (4 Apr, Shift 1)Easy
The velocity of a particle executing simple harmonic motion along $x$-axis is described as $v^2 = 50 - x^2$, where $x$ represents displacement. If the time period of motion is $\dfrac{x}{7}$ s, the value of $x$ is ______.
Step 2. Period. $T=\dfrac{2\pi}{\omega}=2\pi=2\times\dfrac{22}{7}=\dfrac{44}{7}$ s.
Answer. $x=44$.
Solution by Sreeraj P, M.Sc Physics
Q 11-13-043JEE MainJEE Main 2026 (2 Apr, Shift 2)Easy
The equation of motion of a particle is given by $x=a\sin(50t+\pi/3)$ cm. The particle will come to rest at time $t_1$ and it will have zero acceleration at time $t_2$. The $t_1$ and $t_2$ respectively are ______.
Answer: (A) $\dfrac{\pi}{300}$ s, $\dfrac{\pi}{75}$ s
Concept. In $x=a\sin\phi$ with phase $\phi=50t+\dfrac\pi3$, the particle is at rest at an extreme ($\phi=\dfrac\pi2$) and has zero acceleration at the mean ($\phi=\pi$, the next mean position after $t=0$).
Step 1. First rest. $50t_1+\dfrac\pi3=\dfrac\pi2\Rightarrow t_1=\dfrac{\pi}{300}$ s.
Step 2. First zero acceleration. $50t_2+\dfrac\pi3=\pi\Rightarrow t_2=\dfrac{2\pi}{150}=\dfrac{\pi}{75}$ s.
Answer. Option A.
Solution by Sreeraj P, M.Sc Physics
Q 11-13-044JEE MainJEE Main 2026 (28 Jan, Shift 2)Easy
As shown in the figure, a spring is kept in a stretched position with some extension by holding the masses $1$ kg and $0.2$ kg with a separation more than spring natural length and are released. Assuming the horizontal surface to be frictionless, the angular frequency (in SI unit) of the system is :
Concept. Two free blocks joined by a spring oscillate about their fixed centre of mass with $\omega=\sqrt{k/\mu}$, where $\mu=\dfrac{m_1m_2}{m_1+m_2}$.
Step 1. Reduced mass. $\mu=\dfrac{1\times0.2}{1.2}=\dfrac16$ kg.
Q 11-13-045JEE MainJEE Main 2026 (28 Jan, Shift 1)Easy
The displacement of a particle, executing simple harmonic motion with time period $T$, is expressed as $x(t) = A\sin\omega t$, where $A$ is the amplitude. The maximum value of potential energy of this oscillator is found at $t = T/2\beta$. The value of $\beta$ is ______ .
Q 11-13-046JEE MainJEE Main 2026 (24 Jan, Shift 1)Easy
A cylindrical block of mass $M$ and area of cross section $A$ is floating in a liquid of density $\rho$ and with its axis vertical. When depressed a little and released the block starts oscillating. The period of oscillation is ______
Concept. Pushing the block down by $y$ adds buoyancy $\rho Ayg$ upward: a restoring force $-(\rho Ag)y$, so it acts as a spring of constant $\rho Ag$.
Step 1. Period. $T=2\pi\sqrt{\dfrac M{\rho Ag}}$.
Answer. Option C.
Solution by Sreeraj P, M.Sc Physics
Q 11-13-047JEE MainJEE Main 2026 (24 Jan, Shift 1)Easy
A spring of force constant $15\ \text{N/m}$ is cut into two pieces. If the ratio of their length is $1 : 3$, then the force constant of smaller piece is ______ $\text{N/m}$.
Concept. For a spring, $kl=$ constant: a shorter piece is stiffer. The piece with $\tfrac14$ of the length has $4$ times the spring constant.
Step 1. Lengths. $\dfrac L4$ and $\dfrac{3L}{4}$.
Step 2. Smaller piece. $k=4\times15=60$ N m$^{-1}$.
Answer. Option C. (The longer piece has $20$ N m$^{-1}$; in series they give back $15$.)
Solution by Sreeraj P, M.Sc Physics
Q 11-13-048JEE MainJEE Main 2025 (23 Jan, Shift 1)Medium
A light hollow cube of side length $10\ \text{cm}$ and mass $10\ \text{g}$ is floating in water. It is pushed down and released to execute simple harmonic oscillations. The time period of oscillations is $y\pi\times10^{-2}\ \text{s}$, where the value of $y$ is (Acceleration due to gravity $g = 10\ \text{m/s}^2$, density of water $= 10^3\ \text{kg/m}^3$)
Concept. A floating body pushed down by $y$ gets an extra buoyant force $\rho Agy$ upward: SHM with $k=\rho Ag$, so $T=2\pi\sqrt{\dfrac{m}{\rho Ag}}$.
Step 1. Effective $k$. $10^3\times0.01\times10=100$ N m$^{-1}$.
Step 2. Period. $2\pi\sqrt{\dfrac{0.01}{100}}=2\pi\times10^{-2}$ s.
Answer. $y=2$ (Option B).
Solution by Sreeraj P, M.Sc Physics
Q 11-13-049JEE MainJEE Main 2025 (24 Jan, Shift 1)Medium
A particle is executing simple harmonic motion with time period $2\ \text{s}$ and amplitude $1\ \text{cm}$. If $D$ and $d$ are the total distance and displacement covered by the particle in $12.5\ \text{s}$, then $\dfrac{D}{d}$ is
Concept. In each period the particle covers $4A$; start from the mean position (standard reading).
Step 1. Number of periods. $\dfrac{12.5}{2} = 6.25$.
Step 2. Distance. $6 \times 4 + 1 = 25$ cm (the extra quarter takes it from mean to extreme).
Step 3. Displacement. It ends at the extreme: $d = 1$ cm.
Step 4. Ratio. $\dfrac{D}{d} = 25$.
Answer. $25$ (Option D).
Solution by Sreeraj P, M.Sc Physics
Q 11-13-050JEE MainJEE Main 2025 (24 Jan, Shift 2)Medium
A particle oscillates along the $x$-axis according to the law $x(t) = x_0\sin^2\left(\dfrac{t}{2}\right)$, where $x_0 = 1\ \text{m}$. The kinetic energy $(K)$ of the particle as a function of $x$ is correctly represented by the graph
Concept. $\sin^2\dfrac t2=\dfrac12(1-\cos t)$, so $x=\dfrac{x_0}2-\dfrac{x_0}2\cos t$: SHM about $x_0/2=\tfrac12$ m with amplitude $\tfrac12$ m and $\omega=1$.
Step 2. Shape. A downward parabola in $x$: zero at $x=0$ and $x=1$, largest at $x=\tfrac12$.
Answer. Option D.
Solution by Sreeraj P, M.Sc Physics
Q 11-13-051JEE MainJEE Main 2025 (28 Jan, Shift 2)Easy
Given below are two statements. One is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A): Knowing initial position $x_0$ and initial momentum $p_0$ is enough to determine the position and momentum at any time $t$ for a simple harmonic motion with a given angular frequency $\omega$.
Reason (R): The amplitude and phase can be expressed in terms of $x_0$ and $p_0$.
In the light of the above statements, choose the correct answer from the options given below:
Answer: (D) Both (A) and (R) are true and (R) is the correct explanation of (A)
Concept. SHM with known $\omega$ is $x=A\sin(\omega t+\phi)$: only two constants, $A$ and $\phi$.
Step 1. Reason. $x_0=A\sin\phi$ and $p_0=m\omega A\cos\phi$ give $A=\sqrt{x_0^2+\dfrac{p_0^2}{m^2\omega^2}}$ and $\tan\phi=\dfrac{m\omega x_0}{p_0}$. So R is true.
Step 2. Assertion. Once $A$ and $\phi$ are known, $x(t)$ and $p(t)=m\,dx/dt$ are known at all times. A is true, and it is true because of R.
Answer. Both true and R explains A (Option D).
Solution by Sreeraj P, M.Sc Physics
Q 11-13-052JEE MainJEE Main 2025 (29 Jan, Shift 2)Easy
Two bodies A and B of equal mass are suspended from two massless springs of spring constants $k_1$ and $k_2$, respectively. If the bodies oscillate vertically such that their amplitudes are equal, the ratio of the maximum velocity of A to the maximum velocity of B is
Step 1. Ratio. Equal masses and equal amplitudes: $\dfrac{v_A}{v_B}=\sqrt{\dfrac{k_1}{k_2}}$.
Answer. Option B.
Solution by Sreeraj P, M.Sc Physics
Q 11-13-053JEE MainJEE Main 2024 (1 Feb, Shift 2)Easy
A mass $m$ is suspended from a spring of negligible mass and the system oscillates with a frequency $f_1$. The frequency of oscillations if a mass $9m$ is suspended from the same spring is $f_2$. The value of $\dfrac{f_1}{f_2}$ is ______.
Q 11-13-054JEE MainJEE Main 2024 (4 Apr, Shift 2)Easy
In simple harmonic motion, the total mechanical energy of the given system (a particle $P$ of mass $m$ attached to a spring of constant $k$) is $E$. If the mass of the oscillating particle $P$ is doubled, then the new energy of the system for the same amplitude is
Concept. Energy of SHM $E=\tfrac12m\omega^2A^2=\tfrac12kA^2$. With a fixed spring, doubling $m$ halves $\omega^2$, so $m\omega^2=k$ is unchanged.
Step 1. New energy. $\tfrac12kA^2=E$.
Answer. Option A. The mass changes the period, not the energy.
Solution by Sreeraj P, M.Sc Physics
Q 11-13-055JEE MainJEE Main 2024 (4 Apr, Shift 2)Easy
The displacement of a particle executing SHM is given by $x = 10\sin\left(\omega t + \dfrac{\pi}{3}\right)\ \text{m}$. The time period of motion is $3.14\ \text{s}$. The velocity of the particle at $t = 0$ is ______ m/s.
Step 2. Velocity at $t=0$. $v=10\times2\times\cos\dfrac\pi3=20\times\dfrac12=10$ m/s.
Answer. $10$.
Solution by Sreeraj P, M.Sc Physics
Q 11-13-056JEE MainJEE Main 2024 (6 Apr, Shift 1)Easy
A particle is doing simple harmonic motion with an amplitude of $0.06\ \text{m}$ and a time period of $3.14\ \text{s}$. The maximum velocity of the particle is ______ cm/s.
Step 2. Maximum speed. $v_{max}=0.06\times2=0.12$ m/s $=12$ cm/s.
Answer. $12$.
Solution by Sreeraj P, M.Sc Physics
Q 11-13-057JEE MainJEE Main 2024 (27 Jan, Shift 1)Easy
A particle executes simple harmonic motion with an amplitude of $4\ \text{cm}$. At the mean position, velocity of the particle is $10\ \text{cm s}^{-1}$. The distance of the particle from the mean position when its speed becomes $5\ \text{cm s}^{-1}$ is $\sqrt\alpha\ \text{cm}$, where $\alpha =$ ______.
Q 11-13-058JEE MainJEE Main 2024 (27 Jan, Shift 2)Hard
A ball suspended by a thread swings in a vertical plane so that its magnitude of acceleration in the extreme position and lowest position are equal. The angle ($\theta$) of thread deflection in the extreme position will be:
Concept. At the extreme the bob is at rest, so its acceleration is only tangential: $g\sin\theta$. At the lowest point it is only centripetal: $\dfrac{v^2}l=2g(1-\cos\theta)$ by energy conservation.
Q 11-13-059JEE MainJEE Main 2024 (29 Jan, Shift 1)Easy
When the displacement of a simple harmonic oscillator is one third of its amplitude, the ratio of total energy to the kinetic energy is $\dfrac{x}{8}$, where $x =$ ______.
Q 11-13-060JEE MainJEE Main 2024 (29 Jan, Shift 2)Medium
A simple harmonic oscillator has an amplitude $A$ and time period $6\pi$ second. Assuming the oscillation starts from its mean position, the time required by it to travel from $x = A$ to $x = \dfrac{\sqrt3}{2}A$ will be $\dfrac\pi x\ \text{s}$, where $x =$ ______.
Concept. On the reference circle, $x=A$ is at phase $\pi/2$ and $x=\dfrac{\sqrt3}2A$ at phase $\pi/3$: the radius turns through $\pi/6$.
Step 1. Time. $t=\dfrac{\pi/6}{2\pi}T=\dfrac T{12}=\dfrac{6\pi}{12}=\dfrac\pi2$ s.
Answer. $x=2$.
Solution by Sreeraj P, M.Sc Physics
Q 11-13-061JEE MainJEE Main 2024 (8 Apr, Shift 2)Easy
An object of mass $0.2\ \text{kg}$ executes simple harmonic motion along the $x$ axis with frequency of $\left(\dfrac{25}{\pi}\right)\ \text{Hz}$. At the position $x = 0.04\ \text{m}$ the object has kinetic energy $0.5\ \text{J}$ and potential energy $0.4\ \text{J}$. The amplitude of oscillation is ______ cm.
Concept. Total energy $E=K+U=\tfrac12m\omega^2A^2$ is the same at every point. Compute $E$ from the given point and solve for $A$.
Step 1. $\omega$. $2\pi\times\dfrac{25}{\pi}=50$ rad s$^{-1}$.
Step 2. Total energy. $0.5+0.4=0.9$ J.
Step 3. Amplitude. $0.9=\tfrac12\times0.2\times2500\times A^2\Rightarrow A^2=0.0036\Rightarrow A=0.06$ m.
Answer. $6$ cm. (Check: $U=\tfrac12\times0.2\times2500\times0.04^2=0.4$ J, consistent.)
Solution by Sreeraj P, M.Sc Physics
Q 11-13-062JEE MainJEE Main 2024 (9 Apr, Shift 1)Easy
The position, velocity and acceleration of a particle executing simple harmonic motion are found to have magnitudes of $4\ \text{m}$, $2\ \text{m s}^{-1}$ and $16\ \text{m s}^{-2}$ at a certain instant. The amplitude of the motion is $\sqrt x\ \text{m}$, where $x$ is ______.
Q 11-13-063JEE MainJEE Main 2024 (9 Apr, Shift 2)Easy
A particle of mass $0.50\ \text{kg}$ executes simple harmonic motion under the force $F = -50\ (\text{N m}^{-1})\,x$. The time period of oscillation is $\dfrac{x}{35}\ \text{s}$. The value of $x$ is ______. (Given $\pi = \dfrac{22}{7}$)
Step 2. Period. $T=\dfrac{2\pi}{10}=\dfrac{2}{10}\times\dfrac{22}{7}=\dfrac{22}{35}$ s.
Answer. $x=22$.
Solution by Sreeraj P, M.Sc Physics
Q 11-13-064JEE MainJEE Main 2024 (30 Jan, Shift 2)Medium
A simple pendulum is placed at a place where its distance from the earth's surface is equal to the radius of the earth. If the length of the string is $4\ \text{m}$, then the time period of small oscillations will be ______ s. [take $g = \pi^2\ \text{m s}^{-2}$]
Concept. At height $h$ above the surface, $g'=g\dfrac{R^2}{(R+h)^2}$. At $h=R$, $g'=\dfrac g4$.
Step 1. $g'$. $\dfrac{\pi^2}{4}$ m s$^{-2}$.
Step 2. Period. $T=2\pi\sqrt{\dfrac{4}{\pi^2/4}}=2\pi\times\dfrac4\pi=8$ s.
Answer. $8$.
Solution by Sreeraj P, M.Sc Physics
Q 11-13-065JEE MainJEE Main 2024 (31 Jan, Shift 1)Medium
A particle performs simple harmonic motion with amplitude $A$. Its speed is increased to three times at an instant when its displacement is $\dfrac{2A}{3}$. The new amplitude of motion is $\dfrac{nA}{3}$. The value of $n$ is ______.
Concept. Reduce the network: springs that stretch equally are in parallel ($k_1+k_2$); springs that carry the same force are in series ($k_1k_2/(k_1+k_2)$).
Step 1. Right branch. Two parallel springs under the bar: $2k$. In series with the lower spring $k$: $\dfrac{2k\cdot k}{3k}=\dfrac{2k}3$.
Step 2. Whole system. The left spring $k$ and the right branch both move with $M$, so they are in parallel: $k_{eq}=k+\dfrac{2k}3=\dfrac{5k}3$.
Q 11-13-067JEE MainJEE Main 2023 (29 Jan, Shift 2)Easy
A particle of mass $250\ \text{g}$ executes a simple harmonic motion under a periodic force $F=(-25x)\ \text{N}$. The particle attains a maximum speed of $4\ \text{m s}^{-1}$ during its oscillation. The amplitude of the motion is ______ cm.
Step 2. Amplitude. $A=\dfrac{v_{max}}{\omega}=\dfrac{4}{10}=0.4$ m $=40$ cm.
Answer. $40$.
Solution by Sreeraj P, M.Sc Physics
Q 11-13-068JEE MainJEE Main 2023 (30 Jan, Shift 1)Medium
The general displacement of a simple harmonic oscillator is $x=A\sin\omega t$. Let $T$ be its time period. The slope of its potential energy ($U$)–time ($t$) curve will be maximum when $t=\dfrac{T}{\beta}$. The value of $\beta$ is ______.
Answer. $\beta=8$. ($U$ changes fastest midway in time between mean and extreme.)
Solution by Sreeraj P, M.Sc Physics
Q 11-13-069JEE MainJEE Main 2023 (30 Jan, Shift 2)Easy
For a simple harmonic motion in a mass spring system shown, the surface is frictionless. When the mass of the block is $1\ \text{kg}$, the angular frequency is $\omega_1$. When the mass of the block is $2\ \text{kg}$ the angular frequency is $\omega_2$. The ratio $\dfrac{\omega_2}{\omega_1}$ is
Q 11-13-070JEE MainJEE Main 2023 (30 Jan, Shift 2)Easy
The velocity of a particle executing SHM varies with displacement ($x$) as $4v^2=50-x^2$. The time period of oscillations is $\dfrac{x}{7}$ s. The value of $x$ is ______. [Take $\pi=\frac{22}{7}$]
Concept. In SHM $v^2=\omega^2(A^2-x^2)$. Write the given relation in this form and read off $\omega$.
Step 1. Rearrange. $v^2=\dfrac14(50-x^2)$, so $\omega^2=\dfrac14$, $\omega=\dfrac12$ rad s$^{-1}$.
Step 2. Period. $T=\dfrac{2\pi}{\omega}=4\pi=4\times\dfrac{22}{7}=\dfrac{88}{7}$ s.
Answer. $x=88$. (Amplitude $\sqrt{50}$ is not needed.)
Solution by Sreeraj P, M.Sc Physics
Q 11-13-071JEE MainJEE Main 2023 (31 Jan, Shift 1)Easy
The maximum potential energy of a block executing simple harmonic motion is $25\ \text{J}$. $A$ is amplitude of oscillation. At $\dfrac A2$, the kinetic energy of the block is
Concept. Maximum PE $=$ total energy $E=\tfrac12kA^2$. At position $x$: $U=E\dfrac{x^2}{A^2}$ and $K=E-U$.
Step 1. At $A/2$. $U=\dfrac E4$, so $K=\dfrac34E$.
Step 2. Value. $\dfrac34\times25=18.75$ J.
Answer. Option C.
Solution by Sreeraj P, M.Sc Physics
Q 11-13-072JEE MainJEE Main 2023 (31 Jan, Shift 1)Medium
In the figure given below, a block of mass $M=490\ \text{g}$ placed on a frictionless table is connected with two springs having same spring constant ($K=2\ \text{N m}^{-1}$). If the block is horizontally displaced through $X$ m then the number of complete oscillations it will make in $14\pi$ seconds will be ______.
Concept. A block between two springs fixed to walls: a displacement $X$ stretches one and compresses the other, and both forces point back, so $k_{eq}=K+K$.
Step 1. Constant. $k_{eq}=4$ N/m.
Step 2. Period. $T=2\pi\sqrt{\dfrac{0.49}{4}}=2\pi\times0.35=0.7\pi$ s.
Step 3. Oscillations. $\dfrac{14\pi}{0.7\pi}=20$.
Answer. $20$.
Solution by Sreeraj P, M.Sc Physics
Q 11-13-073JEE MainJEE Main 2023 (24 Jan, Shift 1)Easy
A block of mass $2\ \text{kg}$ is attached with two identical springs of spring constant $20\ \text{N m}^{-1}$ each. The block is placed on a frictionless surface and the ends of the springs are attached to rigid supports (see figure). When the mass is displaced from its equilibrium position, it executes a simple harmonic motion. The time period of oscillation is $\dfrac{\pi}{\sqrt X}$ in SI unit. The value of $X$ is ______.
Concept. Block between two springs attached to walls: $k_{eq}=k+k$.
Step 1. Constant. $k_{eq}=40$ N m$^{-1}$.
Step 2. Period. $T=2\pi\sqrt{\dfrac{2}{40}}=\dfrac{2\pi}{\sqrt{20}}=\dfrac{\pi}{\sqrt5}$ s.
Answer. $X=5$.
Solution by Sreeraj P, M.Sc Physics
Q 11-13-074JEE MainJEE Main 2023 (24 Jan, Shift 2)Easy
A mass $m$ attached to free end of a spring executes SHM with a period of $1\ \text{s}$. If the mass is increased by $3\ \text{kg}$ the period of oscillation increases by one second, the value of mass $m$ is ______ kg.
Q 11-13-075JEE MainJEE Main 2023 (25 Jan, Shift 2)Medium
A particle executes simple harmonic motion between $x=-A$ and $x=+A$. If time taken by particle to go from $x=0$ to $\dfrac A2$ is $2$ s, then time taken by particle in going from $x=\dfrac A2$ to $A$ is
Step 1. Check each. (A) True: $|F|\propto x$. (B) True: the minus sign. (C) True: $v=\omega\sqrt{A^2-x^2}$ is largest at $x=0$. (D) False: $|a|=\omega^2|x|$ is largest at the extremes.
Answer. (A), (B) and (C) only (Option C).
Solution by Sreeraj P, M.Sc Physics
Q 11-13-078JEE MainJEE Main 2023 (10 Apr, Shift 2)Easy
For a periodic motion represented by the equation $y=\sin\omega t+\cos\omega t$, the amplitude of the motion is
Concept. $a\sin\omega t+b\cos\omega t=\sqrt{a^2+b^2}\,\sin(\omega t+\alpha)$: one SHM with amplitude $\sqrt{a^2+b^2}$.
Step 1. Amplitude. $\sqrt{1^2+1^2}=\sqrt2$.
Answer. $\sqrt2$ (Option D).
Solution by Sreeraj P, M.Sc Physics
Q 11-13-079JEE MainJEE Main 2023 (10 Apr, Shift 2)Easy
A rectangular block of mass $5\ \text{kg}$ attached to a horizontal spiral spring executes simple harmonic motion of amplitude $1\ \text{m}$ and time period $3.14\ \text{s}$. The maximum force exerted by spring on block is ______ N.
Step 2. Maximum force. $F_{max}=5\times2^2\times1=20$ N.
Answer. $20$.
Solution by Sreeraj P, M.Sc Physics
Q 11-13-080JEE MainJEE Main 2023 (11 Apr, Shift 1)Easy
The variation of kinetic energy (KE) of a particle executing simple harmonic motion with the displacement ($x$) starting from mean position to extreme position ($A$) is given by
Step 1. Shape. From $x=0$ to $x=A$: largest at the mean, falling to zero at $A$ along a downward-opening parabola (flat at $x=0$, steepest at $A$).
Answer. Option C.
Solution by Sreeraj P, M.Sc Physics
Q 11-13-081JEE MainJEE Main 2023 (12 Apr, Shift 1)Easy
A particle is executing simple harmonic motion (SHM). The ratio of potential energy and kinetic energy of the particle when its displacement is half of its amplitude will be
Concept. Total energy minus potential energy is the kinetic energy: $E-U=\tfrac12k(A^2-x^2)$.
Step 1. Shape. A downward parabola, largest at $x=0$ and zero at $x=\pm A$.
Answer. Option B.
Solution by Sreeraj P, M.Sc Physics
Q 11-13-083JEE MainJEE Main 2023 (13 Apr, Shift 1)Easy
At a given point of time the value of displacement of a simple harmonic oscillator is given as $y=A\cos(30^\circ)$. If amplitude is $40\ \text{cm}$ and kinetic energy at that time is $200\ \text{J}$, the value of force constant is $1.0\times10^x\ \text{N m}^{-1}$. The value of $x$ is ______.
Concept. $K=\tfrac12k(A^2-y^2)$ at displacement $y$.
Step 1. Displacement. $y=0.4\cos30^\circ=0.2\sqrt3$ m, so $y^2=0.12$ m$^2$.
Step 2. Solve for $k$. $200=\tfrac12k(0.16-0.12)=0.02k\Rightarrow k=10^4$ N m$^{-1}$.
Answer. $x=4$.
Solution by Sreeraj P, M.Sc Physics
Q 11-13-084JEE MainJEE Main 2023 (6 Apr, Shift 1)Easy
A mass $m$ is attached to two springs as shown in figure. The spring constants of two springs are $K_1$ and $K_2$. For the frictionless surface, the time period of oscillation of mass $m$ is
Concept. When the block moves by $x$, one spring is stretched and the other compressed by $x$; both forces oppose the motion: $k_{eq}=K_1+K_2$.
Step 1. Period. $T=2\pi\sqrt{\dfrac m{K_1+K_2}}$.
Answer. Option A.
Solution by Sreeraj P, M.Sc Physics
Q 11-13-085JEE MainJEE Main 2023 (6 Apr, Shift 2)Medium
A simple pendulum with length $100\ \text{cm}$ and bob of mass $250\ \text{g}$ is executing S.H.M. of amplitude $10\ \text{cm}$. The maximum tension in the string is found to be $\dfrac x{40}$ N. The value of $x$ is ______.
Concept. Tension is largest at the lowest point, where it supplies the centripetal force: $T_{\max}=mg+\dfrac{mv_{\max}^2}{L}$. For SHM of linear amplitude $A$, $v_{\max}=A\omega=A\sqrt{g/L}$.
Step 1. $v_{\max}^2$. $A^2\dfrac gL=0.01g$ (with $L=1$ m).
Step 2. Tension. $T_{\max}=mg(1+0.01)=0.25\times9.8\times1.01=2.475$ N.
Answer. $99$ (with $g=9.8$ m s$^{-2}$; $g=10$ would give $101$).
Solution by Sreeraj P, M.Sc Physics
Q 11-13-086JEE MainJEE Main 2023 (8 Apr, Shift 2)Medium
For particle $P$ revolving round the centre $O$ with radius of circular path $r$ and angular velocity $\omega$, as shown in below figure, the projection of $OP$ on the $x$-axis at time $t$ is
Concept. SHM is the projection of uniform circular motion. If the radius $OP$ makes angle $\theta$ with the $x$-axis, its projection on the $x$-axis is $r\cos\theta$.
Step 1. Angle at time $t$. At $t=0$, $OP$ is at $30^\circ=\pi/6$ above the $x$-axis and turns anticlockwise at $\omega$: $\theta=\omega t+\dfrac\pi6$.
Q 11-13-087JEE MainJEE Main 2023 (10 Apr, Shift 1)Easy
A particle executes S.H.M. of amplitude $A$ along $x$-axis. At $t=0$, the position of the particle is $x=\dfrac A2$ and it moves along positive $x$-axis. The displacement of particle in time $t$ is $x=A\sin(\omega t+\delta)$, then the value of $\delta$ will be
Concept. The initial phase is fixed by the starting position and the direction of motion. Put $t=0$ in $x=A\sin(\omega t+\delta)$ and in $v=A\omega\cos(\omega t+\delta)$.
Step 1. Position. $\dfrac A2=A\sin\delta\Rightarrow\sin\delta=\dfrac12$, so $\delta=\dfrac\pi6$ or $\dfrac{5\pi}6$.
Step 2. Direction. Moving along $+x$ needs $v>0$, i.e. $\cos\delta>0$, which picks $\delta=\dfrac\pi6$.
Answer. Option B. ($\dfrac{5\pi}{6}$ would be the same point while moving towards the mean.)
Solution by Sreeraj P, M.Sc Physics
Q 11-13-088JEE MainJEE Main 2023 (1 Feb, Shift 1)Easy
The amplitude of a particle executing SHM is $3$ cm. The displacement at which its kinetic energy will be $25\%$ more than the potential energy is ______ cm.
Q 11-13-089JEE MainJEE Main 2023 (1 Feb, Shift 2)Medium
A block is fastened to a horizontal spring. The block is pulled to a distance $x=10$ cm from its equilibrium position (at $x=0$) on a frictionless surface from rest. The energy of the block at $x=5$ cm is $0.25$ J. The spring constant of the spring is ______ $\text{N m}^{-1}$.
Concept. Released from rest at $10$ cm, so $A=0.10$ m. The "energy of the block" is its kinetic energy (the block has no PE; the spring has it): $K=\tfrac12k(A^2-x^2)$.
Concept. $T=2\pi\sqrt{L/g}$ gives $T^2=\dfrac{4\pi^2}{g}L$: $T^2$ is proportional to $L$.
Step 1. Graph. A straight line through the origin with slope $4\pi^2/g$, whichever quantity is on which axis.
Answer. Option C.
Solution by Sreeraj P, M.Sc Physics
Q 11-13-091JEE MainJEE Main 2022 (24 Jun, Shift 2)Medium
Two massless springs with spring constants $2k$ and $9k$ carry $50\ \text{g}$ and $100\ \text{g}$ masses at their free ends. These two masses oscillate vertically such that their maximum velocities are equal. Then, the ratio of their respective amplitudes will be:
Q 11-13-092JEE MainJEE Main 2022 (26 Jun, Shift 1)Easy
The time period of a simple pendulum in a stationary lift is $T$. If the lift accelerates with $\dfrac g6$ vertically upwards then the time period will be (where $g$ = acceleration due to gravity)
Concept. In an accelerating lift, use the effective gravity $g_{\text{eff}}=g+a$ for upward acceleration (pseudo-force downward). $T=2\pi\sqrt{L/g_{\text{eff}}}$.
Q 11-13-093JEE MainJEE Main 2022 (25 Jul, Shift 1)Medium
In figure (A), mass $2m$ is fixed on mass $m$ which is attached to two springs of spring constant $k$. In figure (B), mass $m$ is attached to two springs of spring constant $k$ and $2k$. If mass $m$ in (A) and (B) are displaced by distance $x$ horizontally and then released, then the time periods $T_1$ and $T_2$ corresponding to (A) and (B) respectively follow the relation
Step 1. Rearrange. $\dfrac{x^2}{A^2}+\dfrac{v^2}{A^2\omega^2}=1$: an ellipse with semi-axes $A$ and $A\omega$.
Answer. Elliptical (Option B).
Solution by Sreeraj P, M.Sc Physics
Q 11-13-095JEE MainJEE Main 2022 (26 Jul, Shift 2)Medium
As per the given figures, two springs of spring constants $k$ and $2k$ are connected to mass $m$. If the period of oscillation in figure (a) is $3\ \text{s}$, then the period of oscillation in figure (b) will be $\sqrt x\ \text{s}$. The value of $x$ is ______.
Step 3. Ratio. $\dfrac{T_b}{T_a}=\sqrt{\dfrac{k_a}{k_b}}=\sqrt{\dfrac{2/3}{3}}=\dfrac{\sqrt2}3$, so $T_b=\sqrt2$ s.
Answer. $x=2$.
Solution by Sreeraj P, M.Sc Physics
Q 11-13-096JEE MainJEE Main 2022 (27 Jun, Shift 1)Easy
The displacement of simple harmonic oscillator after $3$ seconds starting from its mean position is equal to half of its amplitude. The time period of harmonic motion is
Step 2. Time. $\dfrac{2\pi}{T}\times3=\dfrac\pi6\Rightarrow T=36$ s.
Answer. Option C. Useful fact: mean to $A/2$ takes $T/12$.
Solution by Sreeraj P, M.Sc Physics
Q 11-13-097JEE MainJEE Main 2022 (27 Jun, Shift 2)Easy
The equation of a particle executing simple harmonic motion is given by $x = \sin\pi\left(t + \dfrac13\right)$ m. At $t = 1$ s, the speed of particle will be (Given: $\pi = 3.14$)
Q 11-13-098JEE MainJEE Main 2022 (27 Jun, Shift 2)Easy
A particle executes simple harmonic motion. Its amplitude is $8$ cm and time period is $6$ s. The time it will take to travel from its position of maximum displacement to the point corresponding to half of its amplitude, is ______ s.
Concept. From an extreme ($x=A$, phase $\pi/2$) to $x=A/2$ (phase $\pi/6$) the radius of the reference circle turns through $\pi/3$.
Step 1. Time. $t=\dfrac{\pi/3}{2\pi}T=\dfrac T6=\dfrac66=1$ s.
Answer. $1$.
Solution by Sreeraj P, M.Sc Physics
Q 11-13-099JEE MainJEE Main 2022 (27 Jul, Shift 1)Hard
A mass $0.9$ kg, attached to a horizontal spring, executes SHM with an amplitude $A_1$. When this mass passes through its mean position, then a smaller mass of $124$ g is placed over it and both masses move together with amplitude $A_2$. If the ratio $\dfrac{A_1}{A_2}$ is $\dfrac{\alpha}{\alpha - 1}$, then the value of $\alpha$ will be ______ .
Concept. The small mass is placed at the mean position, where all energy is kinetic. Horizontal momentum is conserved during the placing (it is like a perfectly inelastic collision), and the new amplitude is $A=\dfrac{v}{\omega}$ with the new $\omega$.
Step 1. New speed. $v'=\dfrac{M}{M+m}v$.
Step 2. New $\omega$. $\omega'=\omega\sqrt{\dfrac{M}{M+m}}$.
Step 3. Ratio. $\dfrac{A_2}{A_1}=\dfrac{v'/\omega'}{v/\omega}=\sqrt{\dfrac M{M+m}}$, so $\dfrac{A_1}{A_2}=\sqrt{\dfrac{1.024}{0.9}}=\dfrac{16}{15}$.
Q 11-13-100JEE MainJEE Main 2021 (27 Jul, Shift 1)Easy
A particle starts executing simple harmonic motion (SHM) of amplitude $a$ and total energy $E$. At any instant, its kinetic energy is $\dfrac{3E}{4}$, then its displacement $y$ is given by:
Q 11-13-101JEE MainJEE Main 2021 (27 Jul, Shift 2)Easy
An object of mass $0.5$ kg is executing simple harmonic motion. Its amplitude is $5$ cm and time period $(T)$ is $0.2$ s. What will be the potential energy of the object at an instant $t = \dfrac T4$ s starting from mean position? Assume that the initial phase of the oscillation is zero.
Concept. Starting at the mean with zero phase, $x=A\sin\omega t$; at $t=T/4$ the body is at the extreme, so all its energy is potential: $U=\tfrac12m\omega^2A^2$.
Step 2. Energy. $U=\tfrac12\times0.5\times(10\pi)^2\times(0.05)^2=0.25\times987\times0.0025\approx0.62$ J.
Answer. $0.62$ J (Option A).
Solution by Sreeraj P, M.Sc Physics
Q 11-13-102JEE MainJEE Main 2021 (27 Jul, Shift 2)Easy
A particle executes simple harmonic motion represented by displacement function as $x(t) = A\sin(\omega t + \phi)$. If the position and velocity of the particle at $t = 0$ s are $2$ cm and $2\omega\ \text{cm s}^{-1}$ respectively, then its amplitude is $x\sqrt2$ cm where the value of $x$ is ______.
Q 11-13-103JEE MainJEE Main 2021 (31 Aug, Shift 1)Easy
A particle of mass $1$ kg is hanging from a spring of force constant $100\ \text{N m}^{-1}$. The mass is pulled slightly downward and released so that it executes free simple harmonic motion with time period $T$. The time when the kinetic energy and potential energy of the system will become equal, is $\dfrac Tn$. The value of $n$ is ______.
Concept. $K=U$ at $x=\dfrac{A}{\sqrt2}$. The mass is released from an extreme, so $x=A\cos\omega t$ (measured from the equilibrium position; for a vertical spring $U$ here is the SHM potential energy about that point).
Q 11-13-104JEE MainJEE Main 2021 (31 Aug, Shift 2)Medium
A bob of mass $m$ suspended by a thread of length $\ell$ undergoes simple harmonic oscillations with time period $T$. If the bob is immersed in a liquid that has density $\dfrac14$ times that of the bob and the length of the thread is increased by $\dfrac13$rd of the original length, then the time period of the simple harmonic oscillations will be:
Concept. In a liquid, buoyancy reduces the effective weight: $g_{\text{eff}}=g\left(1-\dfrac{\rho_l}{\rho_b}\right)$. Combine with the change of length in $T\propto\sqrt{L/g_{\text{eff}}}$.
Concept. $K$ and $U$ swap back and forth while their sum stays fixed; over one full period each averages to $\dfrac E2$.
Step 1. (a) False: $K=U$ only at $x=\pm\dfrac A{\sqrt2}$.
Step 2. (b) False: over an arbitrary interval the averages differ (e.g. a short interval near the mean).
Step 3. (c), (d) True.
Answer. Option A.
Solution by Sreeraj P, M.Sc Physics
Q 11-13-106JEE MainJEE Main 2021 (24 Feb, Shift 1)Medium
In the given figure, a mass $M$ is attached to a horizontal spring which is fixed on one side to a rigid support. The spring constant of the spring is $k$. The mass oscillates on a frictionless surface with time period $T$ and amplitude $A$. When the mass is in equilibrium position, as shown in the figure, another mass $m$ is gently fixed upon it. The new amplitude of oscillation will be:
Concept. At the mean position the mass $m$ is added gently: horizontal momentum is conserved at that instant, and the new system has $\omega'=\sqrt{k/(M+m)}$.
Step 3. New amplitude. $A'=\dfrac{v'}{\omega'}=A\dfrac{\sqrt{kM}}{M+m}\sqrt{\dfrac{M+m}k}=A\sqrt{\dfrac M{M+m}}$.
Answer. Option D.
Solution by Sreeraj P, M.Sc Physics
Q 11-13-107JEE MainJEE Main 2021 (24 Feb, Shift 2)Easy
In the given figure, a body of mass $M$ is held between two massless springs, on a smooth inclined plane. The free ends of the springs are attached to firm supports. If each spring has spring constant $k$, the frequency of oscillation of given body is:
Concept. Gravity along the incline is a constant force: it only moves the mean position. Displacing the block by $x$ along the incline changes both spring forces by $kx$ against the motion.
Step 1. Constant. $k_{eq}=2k$.
Step 2. Frequency. $f=\dfrac1{2\pi}\sqrt{\dfrac{2k}M}$, independent of $\alpha$.
Answer. Option D.
Solution by Sreeraj P, M.Sc Physics
Q 11-13-108JEE MainJEE Main 2021 (24 Feb, Shift 2)Easy
When a particle executes SHM, the nature of graphical representation of velocity as a function of displacement is:
Concept. $T=2\pi\sqrt{L/g}$, so $g=\dfrac{4\pi^2L}{T^2}$.
Step 1. Substitute. $g=\dfrac{4\pi^2\times2}{2^2}=2\pi^2$ m s$^{-2}$.
Answer. $2\pi^2$ m s$^{-2}$ (Option A).
Solution by Sreeraj P, M.Sc Physics
Q 11-13-111JEE MainJEE Main 2021 (20 Jul, Shift 2)Medium
A particle is making simple harmonic motion along the $X$-axis. If at a distances $x_1$ and $x_2$ from the mean position the velocities of the particle are $v_1$ and $v_2$, respectively. The time period of its oscillation is given as:
Q 11-13-112JEE MainJEE Main 2021 (22 Jul, Shift 1)Medium
The motion of a mass on a spring, with spring constant $K$ is as shown in figure.
The equation of motion is given by, $x(t) = A\sin\omega t + B\cos\omega t$ with $\omega = \sqrt{\dfrac{K}{m}}$.
Suppose that at time $t = 0$, the position of mass is $x(0)$ and velocity $v(0)$, then its displacement can also be represented as $x(t) = C\cos(\omega t - \phi)$, where $C$ and $\phi$ are
Q 11-13-113JEE MainJEE Main 2021 (22 Jul, Shift 1)Easy
$T_0$ is the time period of a simple pendulum at a place. If the length of the pendulum is reduced to $\dfrac{1}{16}$ times of its initial value, the modified time period is
Step 1. Rearrange. $\dfrac{x^2}{A^2}+\dfrac{v^2}{A^2\omega^2}=1$: an ellipse (a circle only in the special case $\omega=1$).
Answer. An ellipse (Option D).
Solution by Sreeraj P, M.Sc Physics
Q 11-13-117JEE MainJEE Main 2021 (26 Feb, Shift 2)Easy
A particle executes S.H.M. with amplitude $A$ and time period $T$. The displacement of the particle when its speed is half of maximum speed is $\frac{\sqrt{x}A}{2}$. The value of $x$ is
Q 11-13-118JEE MainJEE Main 2021 (26 Feb, Shift 2)Medium
Time period of a simple pendulum is $T$. The time taken to complete $\frac{5}{8}$ oscillations starting from mean position is $\frac{\alpha}{12}T$. The value of $\alpha$ is ______.
Answer. $\alpha=7$. (Not $\tfrac58T$: time is not proportional to distance in SHM.)
Solution by Sreeraj P, M.Sc Physics
Q 11-13-119JEE MainJEE Main 2021 (25 Jul, Shift 2)Easy
In a simple harmonic oscillation, what fraction of total mechanical energy is in the form of kinetic energy, when the particle is midway between mean and extreme position.
Concept. $T\propto\sqrt L$, so a small change gives $\dfrac{\Delta T}{T}=\dfrac12\dfrac{\Delta L}{L}$. A slower pendulum makes the clock lose that fraction of every second.
Step 2. Per day. $86400\times5\times10^{-4}=43.2$ s.
Answer. Option A (the clock loses $43.2$ s).
Solution by Sreeraj P, M.Sc Physics
Q 11-13-121JEE MainJEE Main 2021 (26 Aug, Shift 2)Easy
Two simple harmonic motions are represented by the equations $x_1 = 5\sin\left(2\pi t + \frac{\pi}{4}\right)$ and $x_2 = 5\sqrt{2}(\sin2\pi t + \cos2\pi t)$. The amplitude of the second motion is ______ times the amplitude in the first motion.
Concept. Rewrite each motion as a single sine; its coefficient is the amplitude.
Step 1. First motion. $A_1=5$.
Step 2. Second motion. $\sin2\pi t+\cos2\pi t=\sqrt2\sin\left(2\pi t+\dfrac\pi4\right)$, so $x_2=5\sqrt2\times\sqrt2\sin\left(2\pi t+\dfrac\pi4\right)$ and $A_2=10$.
Step 3. Ratio. $A_2/A_1=2$.
Answer. $2$.
Solution by Sreeraj P, M.Sc Physics
Q 11-13-122JEE MainJEE Main 2021 (27 Aug, Shift 1)Medium
The variation of displacement with time of a particle executing free simple harmonic motion is shown in the figure. The potential energy $U(x)$ versus time $(t)$ plot of the particle is correctly shown in figure:
Concept. $U=\tfrac12kx^2$: never negative, zero whenever $x=0$ and largest at each peak or trough of $x$.
Step 1. Read the graph. $x$ is zero at $O$, $A$, $B$ and $C$, with one hump (up or down) between each pair.
Step 2. $U$–$t$. $U$ is zero at $O$, $A$, $B$, $C$ and has a positive hump between each pair: three humps, all above the axis.
Answer. Option D.
Solution by Sreeraj P, M.Sc Physics
Q 11-13-123JEE MainJEE Main 2021 (27 Aug, Shift 2)Easy
Two simple harmonic motions are represented by the equations $y_1 = 10\sin\left(3\pi t + \frac{\pi}{3}\right)$; $y_2 = 5\left(\sin3\pi t + \sqrt{3}\cos3\pi t\right)$. Ratio of amplitude of $y_1$ to $y_2 = x : 1$. The value of $x$ is
Concept. $a\sin\theta+b\cos\theta$ has amplitude $\sqrt{a^2+b^2}$.
Step 1. Second motion. $y_2=5(\sin3\pi t+\sqrt3\cos3\pi t)$: amplitude $5\sqrt{1+3}=10$. (In fact $y_2=10\sin(3\pi t+\pi/3)=y_1$.)
Step 2. Ratio. $10:10=1:1$.
Answer. $x=1$.
Solution by Sreeraj P, M.Sc Physics
Q 11-13-124JEE MainJEE Main 2020 (5 Sep, Shift 2)Medium
A ring is hung on a nail. It can oscillate, without slipping or sliding, (i) in its plane with a time period $T_1$ and (ii) back and forth in a direction perpendicular to its plane, with a period $T_2$. The ratio $\dfrac{T_1}{T_2}$ will be:
Concept. A rigid body pivoted at a distance $d$ from its centre of mass is a physical pendulum: $T=2\pi\sqrt{\dfrac{I}{Mgd}}$, with $I$ about the pivot. Here $d=R$; only $I$ differs for the two directions.
Step 1. In its plane. Axis perpendicular to the ring through the rim: $I=MR^2+MR^2=2MR^2$, so $T_1=2\pi\sqrt{\dfrac{2R}{g}}$.
Step 2. Perpendicular to its plane. Axis is a tangent in the plane of the ring: $I=\tfrac12MR^2+MR^2=\tfrac32MR^2$, so $T_2=2\pi\sqrt{\dfrac{3R}{2g}}$.
Q 11-13-125JEE MainJEE Main 2020 (6 Sep, Shift 1)Easy
An object of mass $m$ is suspended at the end of a massless wire of length $L$ and area of cross-section $A$. The Young's modulus of the material of the wire is $Y$. If the mass is pulled down slightly, its frequency of oscillation along the vertical direction is:
Q 11-13-126JEE MainJEE Main 2020 (6 Sep, Shift 2)Medium
When a particle of mass $m$ is attached to a vertical spring of spring constant $k$ and released, its motion is described by $y(t) = y_0\sin^2\omega t$, where $y$ is measured from the lower end of the unstretched spring. Then $\omega$ is:
Concept. Rewrite $\sin^2\omega t=\tfrac12(1-\cos2\omega t)$: this is SHM with angular frequency $2\omega$ about $y=\tfrac{y_0}2$. That centre is the equilibrium point, where the spring stretch is $\dfrac{mg}{k}$.
Answer. Option C. Trap: the oscillation frequency is $2\omega$, not $\omega$.
Solution by Sreeraj P, M.Sc Physics
Q 11-13-127JEE MainJEE Main 2020 (2 Sep, Shift 2)Medium
The displacement-time graph of a particle executing SHM is given in the figure (sketch is schematic and not to scale). Which of the following statements is/are true for this motion?
(A) The force is zero at $t = \dfrac{3T}{4}$
(B) The magnitude of acceleration is maximum at $t = T$
(C) The speed is maximum at $t = \dfrac{T}{4}$
(D) The P.E. is equal to K.E. of the oscillation at $t = \dfrac{T}{2}$
Concept. The graph is $x=A\cos(2\pi t/T)$. Force and acceleration are $\propto-x$; speed is the magnitude of the slope; $U\propto x^2$.
Step 1. (A) At $t=3T/4$, $x=0$: force zero. True.
Step 2. (B) At $t=T$, $x=A$ (extreme): $|a|$ maximum. True.
Step 3. (C) At $t=T/4$, $x=0$ (mean): speed maximum. True.
Step 4. (D) At $t=T/2$, $x=-A$: all energy is potential, $K=0$. False.
Answer. (A), (B) and (C) (Option A).
Solution by Sreeraj P, M.Sc Physics
Q 11-13-129JEE MainJEE Main 2019 (9 Jan, Shift 2)Easy
A particle is executing simple harmonic motion (SHM) of amplitude $A$, along the $x$-axis, about $x = 0$. When its potential energy (PE) equals kinetic energy (KE), the position of the particle will be:
Q 11-13-130JEE MainJEE Main 2019 (9 Jan, Shift 2)Medium
A rod of mass $M$ and length $2L$ is suspended at its middle by a wire. It exhibits torsional oscillations. If two masses, each of mass $m$, are attached at a distance $L/2$ from its centre on both sides, it reduces the oscillation frequency by $20\%$. The value of ratio $m/M$ is close to
Q 11-13-131JEE MainJEE Main 2019 (10 Jan, Shift 2)Medium
A particle executes simple harmonic motion with an amplitude of $5\ \text{cm}$. When the particle is at $4\ \text{cm}$ from the mean position, the magnitude of its velocity in SI units is equal to that of its acceleration. Then, its periodic time in seconds is:
Step 3. Equate. $0.03\omega=0.04\omega^2\Rightarrow\omega=\dfrac34$ s$^{-1}$, $T=\dfrac{2\pi}{\omega}=\dfrac{8\pi}{3}$ s.
Answer. Option A.
Solution by Sreeraj P, M.Sc Physics
Q 11-13-132JEE MainJEE Main 2019 (10 Jan, Shift 2)Medium
A cylindrical plastic bottle of negligible mass is filled with $310\ \text{ml}$ of water and left floating in a pond with still water. If pressed downward slightly and released, it starts performing simple harmonic motion at angular frequency $\omega$. If the radius of the bottle is $2.5\ \text{cm}$ then $\omega$ is close to: (density of water $= 10^3\ \text{kg/m}^3$)
Concept. A floating cylinder pushed down by $y$ feels extra buoyancy $\rho Agy$: $k=\rho Ag$ and $\omega=\sqrt{\rho Ag/m}$.
Step 1. Mass and area. $m=0.31$ kg (310 ml of water); $A=\pi(0.025)^2=1.96\times10^{-3}$ m$^2$.
Step 2. Angular frequency. $\omega^2=\dfrac{1000\times1.96\times10^{-3}\times9.8}{0.31}=62$, so $\omega\approx7.9$ rad s$^{-1}$.
Answer. $7.9$ rad s$^{-1}$ (Option C).
Solution by Sreeraj P, M.Sc Physics
Q 11-13-133JEE MainJEE Main 2019 (10 Apr, Shift 1)Easy
The displacement of a damped harmonic oscillator is given by $x(t) = e^{-0.1t}\cos(10\pi t + \varphi)$. Here $t$ is in seconds. The time taken for its amplitude of vibration to drop to half of its initial value is close to:
Concept. In damped SHM the amplitude decays as $A_0e^{-bt/2m}$. Here the decay factor is given directly as $e^{-0.1t}$, so $\dfrac{b}{2m}=0.1$ s$^{-1}$.
Step 1. Half amplitude. $e^{-0.1t}=\dfrac12\Rightarrow t=\dfrac{\ln2}{0.1}=6.93$ s.
Answer. Option D ($\approx7$ s).
Solution by Sreeraj P, M.Sc Physics
Q 11-13-134JEE MainJEE Main 2019 (10 Apr, Shift 2)Medium
A simple pendulum of length $L$ is placed between the vertical plates of a parallel plate capacitor having a horizontal electric field $E$. Its bob has mass $m$ and charge $q$. The time period of the pendulum is given by:
Concept. A constant extra force adds to gravity as a vector. The electric force $qE$ is horizontal, so $g_{\text{eff}}=\sqrt{g^2+(qE/m)^2}$, and the pendulum swings about the tilted equilibrium.
Q 11-13-135JEE MainJEE Main 2019 (9 Jan, Shift 1)Medium
A block of mass $m$, lying on a smooth horizontal surface, is attached to a spring (of negligible mass) of spring constant $k$. The other end of the spring is fixed, as shown in the figure. The block is initially at rest in its equilibrium position. If now the block is pulled with a constant force $F$, the maximum speed of the block is
Concept. A constant force only shifts the equilibrium: the new mean is at $x_0=\dfrac Fk$. The block starts at rest at $x=0$, which is an extreme, so the amplitude is $\dfrac Fk$.
Step 1. Amplitude. $A=\dfrac Fk$.
Step 2. Maximum speed. $v_{\max}=A\omega=\dfrac Fk\sqrt{\dfrac km}=\dfrac{F}{\sqrt{mk}}$.
Answer. Option A. Check by energy: at $x=F/k$, $Fx-\tfrac12kx^2=\tfrac12mv^2$ gives the same.
Solution by Sreeraj P, M.Sc Physics
Q 11-13-136JEE MainJEE Main 2019 (9 Jan, Shift 1)Hard
Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system (see figure). Because of torsional constant $k$, the restoring torque is $\tau = k\theta$ for angular displacement $\theta$. If the rod is rotated by $\theta_0$ and released, the tension in it when it passes through its mean position will be
Concept. Energy: at the mean the twist energy $\tfrac12k\theta_0^2$ has become rotational KE $\tfrac12I\Omega^2$. The rod then supplies each mass its centripetal force $m\Omega^2r$.
Step 1. Centre of mass. From $m$: $\dfrac{(m/2)l}{3m/2}=\dfrac l3$; from $m/2$: $\dfrac{2l}3$.
Step 2. Moment of inertia. $I=m\left(\dfrac l3\right)^2+\dfrac m2\left(\dfrac{2l}3\right)^2=\dfrac{ml^2}9+\dfrac{2ml^2}9=\dfrac{ml^2}3$.
Step 3. Angular speed at the mean. $\Omega^2=\dfrac{k\theta_0^2}I=\dfrac{3k\theta_0^2}{ml^2}$.
Step 4. Tension. $m\Omega^2\dfrac l3=\dfrac{k\theta_0^2}l$ (the same for $m/2$ at $2l/3$).
Answer. $\dfrac{k\theta_0^2}l$ (Option D).
Solution by Sreeraj P, M.Sc Physics
Q 11-13-137JEE MainJEE Main 2019 (8 Apr, Shift 2)Easy
A damped harmonic oscillator has a frequency of 5 oscillations per second. The amplitude drops to half its value for every 10 oscillations. The time it will take to drop to $\dfrac{1}{1000}$ of the original amplitude is close to
Concept. Exponential decay halves the amplitude in equal times. $10$ oscillations at $5$ per second take $2$ s, so the amplitude halves every $2$ s.
Step 1. Number of halvings. $2^{10}=1024\approx1000$: $10$ halvings.
Step 2. Time. $10\times2=20$ s.
Answer. Option C.
Solution by Sreeraj P, M.Sc Physics
Q 11-13-139JEE MainJEE Main 2019 (11 Jan, Shift 1)Easy
A particle undergoing simple harmonic motion has time dependent displacement given by $x(t) = A\sin\dfrac{\pi t}{90}$. The ratio of kinetic to potential energy of this particle at $t = 210$ s will be
Q 11-13-141JEE MainJEE Main 2019 (11 Jan, Shift 2)Hard
A simple pendulum of length $1$ m is oscillating with an angular frequency $10$ rad/s. The support of the pendulum starts oscillating up and down with a small angular frequency of $1$ rad/s and an amplitude of $10^{-2}$ m. The relative change in the angular frequency of the pendulum is best given by:
Concept. The support's acceleration adds to or subtracts from $g$: $g_{eff}=g\pm a$. Since $\omega=\sqrt{g_{eff}/l}$, a small change gives $\dfrac{\Delta\omega}\omega=\dfrac12\dfrac{\Delta g}g$.
Step 1. Support acceleration. $a=A_s\Omega^2=10^{-2}\times1^2=10^{-2}$ m s$^{-2}$.
Step 2. Change in $\omega$. With $g\approx10$ m s$^{-2}$: $\dfrac{\Delta\omega}{\omega}=\dfrac12\times\dfrac{10^{-2}}{10}=5\times10^{-4}$, so $\Delta\omega=10\times5\times10^{-4}=5\times10^{-3}$ rad/s.
Step 3. Order. Of the options, this is of the order of $10^{-3}$ rad/s.
Answer. $10^{-3}$ rad/s (Option A).
Solution by Sreeraj P, M.Sc Physics
Q 11-13-142JEE MainJEE Main 2019 (12 Jan, Shift 1)Medium
Two light identical springs of spring constant $k$ are attached horizontally at the two ends of a uniform horizontal rod AB of length $l$ and mass $m$. The rod is pivoted at its centre O and can rotate freely in horizontal plane. The other ends of the two springs are fixed to rigid supports as shown in figure. The rod is gently pushed through a small angle and released. The frequency of resulting oscillation is:
Concept. Rotational SHM: $\omega^2=\dfrac{\text{restoring torque per radian}}{I}$.
Step 1. Torque. Turning the rod by $\theta$ moves each end by $\dfrac l2\theta$; each spring pulls back with $k\dfrac l2\theta$ at arm $\dfrac l2$. Total torque $2k\left(\dfrac l2\right)^2\theta=\dfrac{kl^2}2\theta$.
Step 2. Moment of inertia. $I=\dfrac{ml^2}{12}$ about the centre.
Step 3. Frequency. $\omega^2=\dfrac{kl^2/2}{ml^2/12}=\dfrac{6k}m$, so $f=\dfrac1{2\pi}\sqrt{\dfrac{6k}m}$.
Answer. Option C.
Solution by Sreeraj P, M.Sc Physics
Q 11-13-143JEE MainJEE Main 2019 (12 Jan, Shift 2)Easy
A simple harmonic motion is represented by: $$y = 5\left(\sin3\pi t + \sqrt3\cos3\pi t\right)\ \text{cm}$$ The amplitude and time period of the motion are:
Concept. $a\sin\theta+b\cos\theta=\sqrt{a^2+b^2}\,\sin(\theta+\phi)$: a sum of sine and cosine of the same frequency is one SHM.
Step 1. Amplitude. $5\sqrt{1+3}=10$ cm, with $\phi=\dfrac\pi3$.
Step 2. Period. $\omega=3\pi\Rightarrow T=\dfrac{2\pi}{3\pi}=\dfrac23$ s.
Answer. Option B.
Solution by Sreeraj P, M.Sc Physics
Q 11-13-144JEE MainJEE Main 2019 (12 Apr, Shift 2)Easy
A spring whose unstretched length is $l$ has a force constant $k$. The spring is cut into two pieces of unstretched lengths $l_1$ and $l_2$ where, $l_1 = nl_2$ and $n$ is an integer. The ratio $k_1/k_2$ of the corresponding force constants, $k_1$ and $k_2$ will be:
Q 11-13-145JEE MainJEE Main 2018 (15 Apr, Shift 1)Hard
A body of mass $M$ and charge $q$ is connected to a spring of spring constant $k$. It is oscillating along $x$-direction about its equilibrium position in the horizontal plane, taken to be at $x = 0$, with an amplitude $A$. An electric field $E$ is applied along the $x$-direction. Which of the following statements is correct?
Answer: (A) The total energy of the system is $\dfrac12 M\omega^2A^2 + \dfrac12\dfrac{q^2E^2}{k}$
Concept. The field adds a constant force $qE$. The new equilibrium is where $kx_0=qE$, $x_0=\dfrac{qE}k$, and the motion is SHM about it with the same $\omega$.
Step 1. Equilibrium. $x_0=qE/k$, so options B and C are wrong.
Step 2. Energy. About the new mean the oscillation energy is $\tfrac12M\omega^2A^2$. Measured from the old origin, the spring holds an extra $\tfrac12kx_0^2=\dfrac12\dfrac{q^2E^2}k$, which is added.
Q 11-13-146JEE MainJEE Main 2018 (15 Apr, Shift 2)Medium
Two simple harmonic motions, $x(t) = A\sin(at + \delta)$ and $y(t) = B\sin(bt)$, are at right angles. They are combined to form Lissajous figures. Identify the correct match below.
Concept. Eliminate $t$ between $x=A\sin(at+\delta)$ and $y=B\sin bt$. With $a=b$: $\delta=0$ gives $\dfrac xA=\dfrac yB$ (a line); $\delta=\pi/2$ gives $\dfrac{x^2}{A^2}+\dfrac{y^2}{B^2}=1$ (an ellipse, a circle if $A=B$).
Step 1. Check the matches. A: $a=2b$ gives a figure-of-eight curve, not a circle. B: $A=B$, $\delta=\pi/2$ gives a circle, not a line. D: $\delta=0$ gives a line, not a parabola. C: $A\ne B$, $a=b$, $\delta=\pi/2$ gives an ellipse: correct.
Answer. Option C.
Solution by Sreeraj P, M.Sc Physics
Q 11-13-147JEE MainJEE Main 2018 (16 Apr, Shift 1)Hard
An oscillator of mass $M$ is at rest in its equilibrium position in a potential, $V = \dfrac12k(x - X)^2$. A particle of mass $m$ comes from the right with speed $u$ and collides completely inelastic with $M$ and sticks to it. This process repeats every time the oscillator crosses its equilibrium position. The amplitude of oscillations after 13 collisions is: $(M = 10,\ m = 5,\ u = 1,\ k = 1)$
Concept. Each particle sticks at the equilibrium position, so only momentum matters there. Track the momentum collision by collision: the oscillator's momentum reverses every half cycle, and each new particle brings $mu$ towards the left.
Step 1. Collision 1. At rest, so after: momentum $mu$ (to the left).
Step 2. Collision 2. Half a cycle later the oscillator returns moving right with momentum $mu$; the new particle brings $mu$ to the left: total $0$. It stops.
Step 3. Pattern. After every even collision it is at rest; after collision $13$, momentum $mu$ with mass $M+13m=75$.
Q 11-13-148JEE MainJEE Main 2018 (16 Apr, Shift 1)Hard
A particle executes simple harmonic motion and it is located at $x = a$, $b$ and $c$ at time $t_0$, $2t_0$ and $3t_0$ respectively. The frequency of the oscillation is:
Concept. For any SHM $x=A\sin(\omega t+\phi)$, use the identity $\sin(\theta-\alpha)+\sin(\theta+\alpha)=2\sin\theta\cos\alpha$ with $\theta=2\omega t_0+\phi$ and $\alpha=\omega t_0$.
Step 1. Add the first and third. $a+c=2A\sin(2\omega t_0+\phi)\cos\omega t_0=2b\cos\omega t_0$.
A particle is executing simple harmonic motion with a time period $T$. At time $t = 0$, it is at its position of equilibrium. The kinetic energy–time graph of the particle will look like:
Concept. Starting at the mean, $x=A\sin\omega t$ and $K=E\cos^2\omega t=\dfrac E2(1+\cos2\omega t)$: largest at $t=0$, period $T/2$.
Step 1. Zeros and peaks. $K=0$ at $t=T/4,\ 3T/4,\dots$ (extremes); $K=E$ at $t=0,\ T/2,\ T$ (mean).
Step 2. Choose. The graph starting at its maximum and touching zero at $T/4$ and $3T/4$.
Answer. Option A.
Solution by Sreeraj P, M.Sc Physics
Q 11-13-150JEE MainJEE Main 2017 (8 Apr)Easy
The ratio of maximum acceleration to maximum velocity in a simple harmonic motion is $10\ \text{s}^{-1}$. At $t = 0$ the displacement is $5$ m. What is the maximum acceleration? The initial phase is $\dfrac\pi4$.
Concept. $v_{max}=A\omega$ and $a_{max}=A\omega^2$, so $\dfrac{a_{max}}{v_{max}}=\omega$. The initial displacement and phase give $A$.
Step 1. Angular frequency. $\omega=10$ s$^{-1}$.
Step 2. Amplitude. $x(0)=A\sin\dfrac\pi4\Rightarrow5=\dfrac{A}{\sqrt2}\Rightarrow A=5\sqrt2$ m.
Step 3. Maximum acceleration. $a_{max}=\omega^2A=100\times5\sqrt2=500\sqrt2$ m s$^{-2}$.
Answer. $500\sqrt2$ m s$^{-2}$ (Option D).
Solution by Sreeraj P, M.Sc Physics
Q 11-13-151JEE MainJEE Main 2017 (8 Apr)Easy
A $1$ kg block attached to a spring vibrates with a frequency of $1$ Hz on a frictionless horizontal table. Two springs identical to the original spring are attached in parallel to an $8$ kg block placed on the same table. So, the frequency of vibration of the $8$ kg block is:
A block of mass $0.1$ kg is connected to an elastic spring of spring constant $640\ \text{N m}^{-1}$ and oscillates in a damping medium of damping constant $10^{-2}\ \text{kg s}^{-1}$. The system dissipates its energy gradually. The time taken for its mechanical energy of vibration to drop to half of its initial value, is closest to:
Step 2. Half energy. $t=\dfrac{\ln2}{0.1}=6.93$ s.
Answer. Option D ($\approx7$ s). Trap: energy decays twice as fast as amplitude.
Solution by Sreeraj P, M.Sc Physics
Q 11-13-154JEE MainJEE Main 2018 (8 Apr)Easy
A silver atom in a solid oscillates in simple harmonic motion in some direction with a frequency of $10^{12}\ \text{s}^{-1}$. What is the force constant of the bonds connecting one atom with the other? (Mole wt. of silver $= 108\ \text{g mol}^{-1}$ and Avogadro number $= 6.02\times10^{23}$)
Concept. Model each bond as a spring: $k=m\omega^2=m(2\pi f)^2$, with the mass of one atom $m=M/N_A$.
Step 1. Mass of an atom. $m=\dfrac{0.108}{6.02\times10^{23}}=1.79\times10^{-25}$ kg.
Step 2. Force constant. $k=1.79\times10^{-25}\times(2\pi\times10^{12})^2=1.79\times10^{-25}\times3.95\times10^{25}\approx7.1$ N m$^{-1}$.
Answer. $7.1$ N m$^{-1}$ (Option C).
Solution by Sreeraj P, M.Sc Physics
Q 11-13-155NEETNEET 2022Top questionMedium
Two pendulums of length $121$ cm and $100$ cm start vibrating in phase. At some instant, the two are at their mean position in the same phase. The minimum number of vibrations of the shorter pendulum after which the two are again in phase at the mean position is:
where $x$ is displacement and $t$ is time in seconds. The maximum acceleration of the particle is $x \times 10^{-2}\ \text{m s}^{-2}$. The value of $x$ is:
Concept. Two SHMs of the same frequency along the same line add like vectors (phasors): $R^2=A_1^2+A_2^2+2A_1A_2\cos\delta$. The result is SHM with the same $\omega$, so $a_{\max}=\omega^2R$.
Step 1. Resultant amplitude. $R^2=7+28+2(\sqrt7)(2\sqrt7)\tfrac12=35+14=49\Rightarrow R=7$ cm.
Step 2. Maximum acceleration. $\omega^2R=25\times7=175$ cm s$^{-2}$ $=175\times10^{-2}$ m s$^{-2}$.
Answer. Option A.
Solution by Sreeraj P, M.Sc Physics
Q 11-13-157JEE MainJEE Main 2025 (3 Apr, Shift 1)Hard
Two blocks of masses $m$ and $M$ ($M > m$) are placed on a frictionless table as shown in the figure. A massless spring with spring constant $k$ is attached to the lower block. If the system is slightly displaced and released, then ($\mu$ = coefficient of friction between the two blocks):
(A) The time period of small oscillation of the two blocks is $T = 2\pi\sqrt{\dfrac{m + M}{k}}$
(B) The acceleration of the blocks is $a = \dfrac{kx}{M + m}$ ($x$ = displacement of the blocks from the mean position)
(C) The magnitude of the frictional force on the upper block is $\dfrac{m\mu|x|}{M + m}$
(D) The maximum amplitude of the upper block, if it does not slip, is $\dfrac{\mu(M + m)g}{k}$
(E) The maximum frictional force can be $\mu(M + m)g$.
Choose the correct answer from the options given below:
Concept. If the upper block does not slip, the two blocks move as one mass $M+m$ on the spring. Friction is what accelerates the upper block, and it cannot exceed $\mu mg$.
Step 1. (A) and (B). $T=2\pi\sqrt{\dfrac{M+m}{k}}$ and $|a|=\dfrac{k|x|}{M+m}$: true.
Step 2. (C). Friction on $m$ is $ma=\dfrac{mk|x|}{M+m}$, not $\dfrac{m\mu|x|}{M+m}$: false.
Step 3. (D). No slip needs $\dfrac{mkA}{M+m}\le\mu mg\Rightarrow A\le\dfrac{\mu(M+m)g}{k}$: true.
Step 4. (E). Friction on $m$ is at most $\mu mg$, not $\mu(M+m)g$: false.
Answer. Option A.
Solution by Sreeraj P, M.Sc Physics
Q 11-13-158JEE MainJEE Main 2025 (4 Apr, Shift 1)Easy
Two simple pendulums having lengths $l_1$ and $l_2$ with negligible string mass undergo angular displacements $\theta_1$ and $\theta_2$ from their mean positions, respectively. If the angular accelerations of both pendulums are the same, then which expression is correct?
Concept. For a simple pendulum the restoring torque is $mgl\sin\theta\approx mgl\theta$ and $I=ml^2$, so the angular acceleration is $\alpha=\dfrac{g}{l}\theta$.
Q 11-13-159JEE MainJEE Main 2025 (8 Apr, Shift 2)Medium
A block of mass $2\ \text{kg}$ is attached to one end of a massless spring whose other end is fixed at a wall. The spring–mass system moves on a frictionless horizontal table. The spring's natural length is $2\ \text{m}$ and its spring constant is $200\ \text{N/m}$. The block is pushed such that the length of the spring becomes $1\ \text{m}$ and then released. At distance $x\ \text{m}$ ($x < 2$) from the wall, the speed of the block will be:
Concept. Natural length $2$ m, compressed to $1$ m and released from rest: amplitude $A=1$ m about the equilibrium at $x=2$ m. The displacement from equilibrium at distance $x$ from the wall is $2-x$.
Solution by Sreeraj P, M.Sc Physics
