Question

A particle executes a simple harmonic motion according to the equation $4\dfrac{{{d^2}x}}{{d{t^2}}} + 320x = 0$. Its time period of oscillation is
A. $\dfrac{{2\pi }}{{5\sqrt 3 }}s$
B. $\dfrac{\pi }{{3\sqrt 2 }}s$
C. $\dfrac{\pi }{{2\sqrt 5 }}s$
D. $\dfrac{{2\pi }}{{\sqrt 3 }}s$

Answer 1

Hint: The motion in which the restoring force is directly proportional to the displacement of the body to the mean position is called the simple harmonic equation. The direction of the given restoring force is always pointed towards the position of the mean position. To solve the question, use the equation of the simple harmonic equation.

Formula used:
$ \Rightarrow \dfrac{{{d^2}x}}{{d{t^2}}} + {\omega ^2}x = 0$
Where, $\omega $ is the angular frequency.

Complete answer:
An oscillatory motion in which the acceleration of any particle at any position is directly proportional to the displacement of the particle from any mean position is called Simple harmonic motion. It is one of the special cases of oscillatory motions. The mathematical representation of the simple harmonic motion is given as,
$ \Rightarrow \dfrac{{{d^2}x}}{{d{t^2}}} + {\omega ^2}x = 0$
Where, $\omega $is the angular frequency.
The solution to the question. Given a particle is executing a simple harmonic equation $4\dfrac{{{d^2}x}}{{d{t^2}}} + 320x = 0$.
To find the time period.
To solve the question, compare the equations of the simple harmonic motion and the particle that executes the simple harmonic motion.
The standard differential equation of the simple harmonic motion:
$ \Rightarrow \dfrac{{{d^2}x}}{{d{t^2}}} + {\omega ^2}x = 0$ …………..1
The equation is given in the question:
$ \Rightarrow 4\dfrac{{{d^2}x}}{{d{t^2}}} + 320x = 0$
Simplify the equation.
$ \Rightarrow \dfrac{{{d^2}x}}{{d{t^2}}} + 80x = 0$…………………2
Compare 1 and 2 equations.
$ \Rightarrow \dfrac{{{d^2}x}}{{d{t^2}}} + {\omega ^2}x = \dfrac{{{d^2}x}}{{d{t^2}}} + 80x$
On comparison, the angular frequency has the value of $80$.
$ \Rightarrow {\omega ^2} = 80$
To remove the power, take square root on the right-hand side:
$ \Rightarrow \omega = \sqrt {80} $
On taking the square root,
$ \Rightarrow \omega = 4\sqrt 5 $
To calculate the time period, the formula is,
$ \Rightarrow T = \dfrac{{2\pi }}{\omega }$
Where, $\omega $is the angular frequency.
The value of the angular frequency is calculated as $4\sqrt 5 $. Substitute the value,
$ \Rightarrow T = \dfrac{{2\pi }}{{4\sqrt 5 }}$
Simplify the equation,
$ \Rightarrow T = \dfrac{\pi }{{2\sqrt 5 }}s$
Therefore, the time period is calculated as $\dfrac{\pi }{{2\sqrt 5 }}s$.
Hence, the option $\left( C \right)$ is correct.

Note:
All the simple harmonic motions are oscillatory motions and periodic motions. Whereas all period motions and oscillatory motions are not harmonic motions. The special case of the simple harmonic motions is that they oscillate along the straight line between any two extreme points and their restoring force will always be directed towards the direction of the mean position.