Question

An object of mass $0.2kg$ executes SHM along the x-axis with a frequency of\[\dfrac{25}{\pi }Hz\]. At the position $x=0.04m$ the object has kinetic energy $0.5J$ and potential energy$0.4J$. The amplitude of oscillation will be
(A) $0.06m$
(B) $0.04m$
(C) $0.05m$
(D) $0.25m$

Answer 1

Hint : In this question, we have the values of mass, frequency, kinetic and potential energy. We are required to find the amplitude of oscillation and in order to find that we will use the formula of Total energy. So by substituting these values we will get the value of amplitude.

Complete step by step answer:
Given:
 Mass $=0.2kg$
 Frequency $=\dfrac{25}{\pi }Hz$
 Position$(x)=0.04m$
 Kinetic energy $=0.5J$
 Potential energy $=0.04J$
We know that Total Energy of Simple Harmonic Motion is,
 $E=\dfrac{1}{2}m{{\omega }^{2}}{{A}^{2}}.......(1)$
Where,
 $E=$Total energy
 $m=$Mass
 $\omega =$Angular frequency
 $A=$Amplitude
As we know,
 $\omega =2\pi f$
The equation (1) becomes:
 $E=\dfrac{1}{2}m{{(2\pi f)}^{2}}{{A}^{2}}......(2)$
Taking square-root on both side of the equation (2), we get
 $A=\dfrac{1}{2\pi f}\sqrt{\dfrac{2E}{m}}......(3)$
Also,
 $E=K+P$
Where,
 $K=$Kinetic energy
 $P=$ Potential energy
Substituting the above equation in equation (3), we get
 $A=\dfrac{1}{2\pi f}\sqrt{\dfrac{2(K+P)}{m}}$
Putting the value of frequency, mass, kinetic energy and potential energy in above equation, we get
 $A=\dfrac{1}{2\pi (\dfrac{25}{\pi })}\sqrt{\dfrac{2\times (0.5+0.4)}{0.2}}$
On simplifying above equation, we get
 $A=\dfrac{1}{50}\sqrt{\dfrac{2(0.9)}{0.2}}$
 $A=\dfrac{3}{50}$
Therefore,
 $A=0.06$

So the amplitude of oscillation will be 0.06 m. Hence option A. is correct.

Note: In order to solve this kind of question we should have knowledge about oscillation, the energy of simple harmonic motion, and basic math. The total energy that a particle possesses while performing simple harmonic motion is energy in simple harmonic motion. Take a pendulum for example, when it is at its mean position, it is at rest. When it moves towards its extreme position, it is in motion and as soon as it reaches its extreme position, it comes to rest again. Therefore, in order to calculate the energy in simple harmonic motion, we should have the value of kinetic and potential energy that the particle possesses.