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. \[\dfrac{1}{2\pi {{t}_{0}}}{{\cos }^{-1}}\left( \dfrac{a+b}{2c} \right)\]
B. \[\dfrac{1}{2\pi {{t}_{0}}}{{\cos }^{-1}}\left( \dfrac{a+b}{3c} \right)\]
C. \[\dfrac{1}{2\pi {{t}_{0}}}{{\cos }^{-1}}\left( \dfrac{2a+3b}{b} \right)\]
D. \[\dfrac{1}{2\pi {{t}_{0}}}{{\cos }^{-1}}\left( \dfrac{a+c}{2b} \right)\]

Answer 1

Hint: A particle is doing simple harmonic motion (SHM), its displacement is given by \[x=A\sin \omega t\]. First we will find the relation between given displacement and time, then we are able to find the frequency of oscillation which is \[\dfrac{2\pi }{\omega }\].

Formula Used:
Displacement of a particle executing SHM is given by:
\[x=A\sin \omega t\]
Where:
\[x=\] displacement of particle at time t
\[a=\]amplitude or maximum displacement of a particle.
\[\omega =\] angular frequency of SHM
\[\omega =2\pi n\]
Where \[n\] is the frequency of oscillation.

Complete step by step answer:
In the given question particle is located at \[a\] at time \[{{t}_{0}}\], therefore,
\[a=A\sin \omega {{t}_{0}}\] ……..(1)
Particle is located at \[b\] at time \[2{{t}_{0}}\]
\[b=A\sin 2\omega {{t}_{0}}\] ……….(2)
Particle is located at \[c\] at time \[3{{t}_{0}}\]
\[c=A\sin 3\omega {{t}_{0}}\] ……….(3)
Adding equation (1) and (3)
\[\begin{align}
  & a+c=A\sin \omega {{t}_{0}}+A\sin 3\omega {{t}_{0}} \\
 & a+c=A(\sin \omega {{t}_{0}}+\sin 3\omega {{t}_{0}}) \\
\end{align}\]
Applying the trigonometry formula of adding to sine function i.e. \[\sin C+\operatorname{sin}D=2\sin \dfrac{C+D}{2}\cos \dfrac{C-D}{2}\]
\[a+c=2A\sin 2\omega {{t}_{0}}\cos \omega {{t}_{0}}\] ……….(4)
Now, dividing equation (4) with equation (2)
\[\begin{align}
  & \dfrac{a+c}{b}=\dfrac{2A\sin 2\omega {{t}_{0}}\cos \omega {{t}_{0}}}{A\sin 2\omega {{t}_{0}}} \\
 & \dfrac{a+c}{b}=2\cos \omega {{t}_{0}} \\
 & \cos \omega {{t}_{0}}=\dfrac{a+c}{2b} \\
 & \omega {{t}_{0}}={{\cos }^{-1}}\left( \dfrac{a+c}{2b} \right) \\
 & 2\pi n{{t}_{0}}={{\cos }^{-1}}\left( \dfrac{a+c}{2b} \right) \\
 & n=\dfrac{1}{2\pi {{t}_{0}}}{{\cos }^{-1}}\left( \dfrac{a+c}{2b} \right) \\
\end{align}\]
Hence, the option D. is the correct answer.

Additional Information:
There are two types of motion which repeat them. These are oscillatory motion and simple harmonic motion, although they look the same but there is little difference between them.
Oscillatory motion: If a particle moves to and fro about a fixed point in a regular interval of time then motion is known as oscillatory motion.
SHM: If a particle moves to and fro about a fixed point in a regular interval of time such that restoring force acting on the particle is directly proportional to its displacement it is known as simple harmonic motion or SHM.

Note: To determine the velocity of a particle doing SHM just differentiate displacement with respect to time and to determine acceleration of particle differentiate velocity with respect to time.
\[x=A\sin \omega t\]
\[\begin{align}
  & v=A\omega \cos \omega t=\omega \sqrt{{{A}^{2}}-{{x}^{2}}} \\
 & a=-A{{\omega }^{2}}\sin \omega t=-A{{\omega }^{2}}x \\
\end{align}\]
At mean position i.e. at x = 0 velocity is maximum i.e. \[A\omega \] and acceleration is minimum i.e. zero. But at amplitude the velocity of particle is zero (minimum) and acceleration is maximum i.e. \[A{{\omega }^{2}}\]