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:
Answer: (A) $\dfrac{1}{2\pi t_0}\cos^{-1}\left(\dfrac{a + c}{2b}\right)$
Let $x = A\sin(\omega t + \phi)$. Then
$$a + c = A\left[\sin(\omega t_0 + \phi) + \sin(3\omega t_0 + \phi)\right] = 2A\sin(2\omega t_0 + \phi)\cos(\omega t_0) = 2b\cos(\omega t_0)$$
So $\cos(\omega t_0) = \dfrac{a + c}{2b}$ and
$$f = \frac{\omega}{2\pi} = \frac{1}{2\pi t_0}\cos^{-1}\left(\frac{a + c}{2b}\right)$$
Solution by Sreeraj P, M.Sc Physics