Question

A particle executes simple harmonic motion and is located at $ x = a,b $ and $ c $ at times $ {t_{0,}}2{t_0} $ and $ 3{t_0} $ respectively. The frequency of the oscillation is
(A) $ \dfrac{1}{{2\pi {t_0}}}{\cos ^{ - 1}}\left( {\dfrac{{2a + 3c}}{b}} \right) $
(B) $ \dfrac{1}{{2\pi {t_0}}}{\cos ^{ - 1}}\left( {\dfrac{{a + c}}{{2b}}} \right) $
(C) $ \dfrac{1}{{2\pi {t_0}}}{\cos ^{ - 1}}\left( {\dfrac{{a + c}}{{2c}}} \right) $
(D) $ \dfrac{1}{{2\pi {t_0}}}{\cos ^{ - 1}}\left( {\dfrac{{a + 2b}}{{2c}}} \right) $

Answer 1

Hint : to solve this problem we should know about simple harmonic motion. The general SHM equation applies to all simple oscillating motion is,
  $ x = {x_0}\cos (\omega t) $
Here, $ {x_0} $ is the amplitude of the SHM and $ \omega $ is the angular frequency of the SHM.

Complete Step By Step Answer:
A particle executes simple harmonic motion having $ A $ is amplitude of SHM and $ \omega $ is angular frequency of the SHM.
As per location given in question. Equation of simple harmonic motion will be at $ x = a,b $ and $ c $ at given time $ {t_{0,}}2{t_0} $ and $ 3{t_0} $ will be respectively,
  $ a = A\cos \omega {t_0} $ ………………………… $ (1) $
  $ b = A\cos 2\omega {t_0} $ ………………………… $ (2) $
  $ c = A\cos 3\omega {t_0} $ ………………………… $ (3) $
On adding $ (1) $ and $ (3) $ . We get,
  $ a + c = A(\cos \omega {t_0} + \cos 3\omega {t_0}) $
By applying a trigonometric equation.
  $ \Rightarrow a + c = 2A\left( {\cos \left( {\dfrac{{3\omega {t_0} + \omega {t_0}}}{2}} \right)\cos \left( {\dfrac{{3\omega {t_0} - \omega {t_0}}}{2}} \right)} \right) $
  $ \Rightarrow a + c = 2A\cos 2\omega {t_0}\cos \omega {t_0} $
From $ (2) $ we get,
  $ b = A\cos 2\omega {t_0} $
  $ \Rightarrow a + c = 2b\cos \omega {t_0} $
By taking the inverse. We get,
  $ \Rightarrow {\cos ^{ - 1}}\left( {\dfrac{{a + c}}{{2b}}} \right) = \omega {t_0} $
As we know $ \omega = 2\pi f $ . So,
  $ \Rightarrow {\cos ^{ - 1}}\left( {\dfrac{{a + c}}{{2b}}} \right) = 2\pi f{t_0} $
  $ \Rightarrow f = \dfrac{1}{{2\pi {t_0}}}{\cos ^{ - 1}}\left( {\dfrac{{a + c}}{{2b}}} \right) $
Hence, (b) $ \dfrac{1}{{2\pi {t_0}}}{\cos ^{ - 1}}\left( {\dfrac{{a + c}}{{2b}}} \right) $ is correct option.

Note :
In simple harmonic motion particles oscillate about their mean position about which particle is to its to and for motion. In simple harmonic motion distance is directly proportional to acceleration of the particle. The maximum kinetic energy of a particle is at mean position and maximum potential energy is at maximum position and vice-versa.