Question

A particle executes linear simple harmonic motion with an amplitude of 3 cm. When the particle is at 2 cm from the mean position, the magnitude of its velocity is equal to that of its acceleration. What is its time period in seconds?
\[
  {\text{A}}{\text{. }}\dfrac{{\sqrt 5 }}{\pi } \\
  {\text{B}}{\text{. }}\dfrac{{\sqrt 5 }}{{2\pi }} \\
  {\text{C}}{\text{. }}\dfrac{{4\pi }}{{\sqrt 5 }} \\
  {\text{D}}{\text{. }}\dfrac{{2\pi }}{{\sqrt 3 }} \\
 \]

Answer 1

Hint: Here, we will proceed by equating the velocity of the particle undergoing linear simple harmonic motion at 2 cm to the acceleration of the same particle at 2 cm. From this equation, we will find the frequency.

Step By Step Answer:
Formulas Used- ${\text{v}} = \omega \sqrt {{{\text{A}}^2} - {x^2}} $ and ${\text{a}} = {\omega ^2}x$.

Amplitude of linear simple harmonic motion A = 3 cm
Position of the particle ${x_0}$ = 2 cm
Let the required time period be ${t_0}$ seconds

Let the frequency of the given particle undergoing linear simple harmonic motion be $\omega $ rad/sec

As we know that for any particle undergoing simple harmonic motion of amplitude A and frequency $\omega $ rad/sec, the velocity of the particle at any position x is given by

${\text{v}} = \omega \sqrt {{{\text{A}}^2} - {x^2}} {\text{ }} \to {\text{(1)}}$

Also, the acceleration of the particle at any position x is given by

${\text{a}} = {\omega ^2}x{\text{ }} \to {\text{(2)}}$

Using the formula given by equation (1), the velocity of the given particle at ${x_0}$ = 2 cm is given by

$
  {\text{v}} = \omega \sqrt {{{\text{A}}^2} - {{\left( {{x_0}} \right)}^2}} \\
   \Rightarrow {\text{v}} = \omega \sqrt {{{\text{3}}^2} - {{\left( 2 \right)}^2}} \\
   \Rightarrow {\text{v}} = \omega \sqrt {9 - 4} \\
   \Rightarrow {\text{v}} = \omega \sqrt 5 {\text{ }} \to {\text{(3)}} \\
 $

Using the formula given by equation (2), the acceleration of the given particle at ${x_0}$ = 2 cm is given by

$
  {\text{a}} = {\omega ^2}{x_0} \\
   \Rightarrow {\text{a}} = 2{\omega ^2}{\text{ }} \to {\text{(4)}} \\
 $

According to the problem statement it is given that at ${x_0}$ = 2 cm, the magnitude of the velocity is equal to the magnitude of the acceleration i.e., v = a

Using equations (3) and (4) in the equation v = a, we get

$
   \Rightarrow \omega \sqrt 5 = 2{\omega ^2} \\
   \Rightarrow \sqrt 5 = 2\omega \\
   \Rightarrow \omega = \dfrac{{\sqrt 5 }}{2} \\
 $

So, the frequency of the given particle undergoing linear simple harmonic motion is $\dfrac{{\sqrt 5 }}{2}$ rad/sec

Also we know that the frequency of any particle undergoing linear simple harmonic motion is given by

$\omega = \dfrac{{2\pi }}{T}$ where T is the time period

Putting $\omega = \dfrac{{\sqrt 5 }}{2}$ in the above equation, we get

\[
   \Rightarrow \dfrac{{\sqrt 5 }}{2} = \dfrac{{2\pi }}{T} \\
   \Rightarrow T\sqrt 5 = 4\pi \\
   \Rightarrow T = \dfrac{{4\pi }}{{\sqrt 5 }} \\
 \]

Therefore, the time period of the given particle undergoing linear simple harmonic motion is \[\dfrac{{4\pi }}{{\sqrt 5 }}\] seconds.

Hence, option C is correct.

Note: In order to solve this problem, one should know about linear SHM. A body's linear simple harmonic motion is a form of motion in which the restorative force is always acting towards the equilibrium position or mean position and its magnitude is directly proportional to the displacement from the equilibrium position.