Periodic Motion & SHM Basics question from JEE Main 2018 (16 Apr, Shift 1)

Answer: (a)

1. Let $x = A\sin(\omega t + \delta)$, so $a = A\sin(\omega t_0 + \delta)$, $b = A\sin(2\omega t_0 + \delta)$, $c = A\sin(3\omega t_0 + \delta)$.
2. $a + c = 2A\sin(2\omega t_0 + \delta)\cos(\omega t_0) = 2b\cos\omega t_0$
3. $\cos\omega t_0 = \frac{a+c}{2b} \Rightarrow \omega = \frac{1}{t_0}\cos^{-1}\left(\frac{a+c}{2b}\right)$
4. $f = \frac{\omega}{2\pi} = \frac{1}{2\pi t_0}\cos^{-1}\left(\frac{a+c}{2b}\right)$
5. So the answer is (a).