Oscillations: NEET and JEE questions with solutions

Answer: (D) $1/4$

For a cork of mass $m$ and cross-section area $A$, pushing it down by $y$ adds a buoyant restoring force $\rho A g y$. So
$$k = \rho A g, \qquad T = 2\pi\sqrt{\frac{m}{\rho A g}} \;\Rightarrow\; T \propto \frac{1}{\sqrt{\rho}}$$

$$\frac{T_2}{T_1} = \sqrt{\frac{\rho_1}{\rho_2}} = 2 \;\Rightarrow\; \frac{\rho_2}{\rho_1} = \frac{1}{4}$$

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $k = m$

Energy conservation: $\dfrac{1}{2}mv^2 + \dfrac{1}{2}kx^2 = \dfrac{1}{2}kA^2$, which gives
$$x^2 + \frac{m}{k}v^2 = A^2$$

This is an ellipse in general. It is a circle when the coefficients of $x^2$ and $v^2$ are equal:
$$\frac{m}{k} = 1 \;\Rightarrow\; k = m$$

(numerically, in SI units; this is the case $\omega = 1\ \text{rad s}^{-1}$).

Solution by Sreeraj P, M.Sc Physics

Answer: (B) see figure

Frequency: $\omega = \sqrt{k/m}$. As sand leaks out, $m$ decreases, so $\omega$ increases with time.

Amplitude: the sand leaves vertically, with the same horizontal velocity as the box, so it exerts no horizontal force on the box. The box obeys $m(t)\ddot{x} = -kx$, i.e. $\ddot{x} + \omega(t)^2x = 0$ with $\omega$ changing slowly.

For such a slowly changing frequency, the quantity $\omega A^2$ stays constant (adiabatic invariant), so $A \propto \omega^{-1/2}$. As $\omega$ increases, the amplitude decreases.

(Energy view: each grain of sand carries away its own kinetic energy, so the mechanical energy of the box, $\tfrac{1}{2}kA^2$, decreases.)

So $\omega(t)$ increases and $A(t)$ decreases, as in option (2).

Solution by Sreeraj P, M.Sc Physics

Answer: (D) $\sqrt{\dfrac{k_1}{k_2}}$

$v_{max} = A\omega = A\sqrt{\dfrac{k}{m}}$. With equal masses and equal $v_{max}$:
$$A_P\sqrt{k_1} = A_Q\sqrt{k_2} \;\Rightarrow\; \frac{A_Q}{A_P} = \sqrt{\frac{k_1}{k_2}}$$

Solution by Sreeraj P, M.Sc Physics