Question

The total energy of the particle executing simple harmonic motion of amplitude A is 100J. At a distance of 0.707A from the mean position, its kinetic energy is –
A) 25J
B) 50J
C) 100J
D) 12.5J
E) 70J

Answer 1

Hint: We have to find the relation between the energy in a simple harmonic motion and the amplitude of the oscillation. We know that the position of the particle at an instant is sine function whose first and second derivatives are the velocity and the acceleration respectively.

Complete answer:
 Let us consider the given situation where a particle is undergoing a simple harmonic motion. The particle position at an instant is given by –
\[x=A\sin \omega t\]
The velocity of the particle and its acceleration is given by –
\[\begin{align}
  & v=A\omega \cos \omega t \\
 & a=-A{{\omega }^{2}}\sin \omega t \\
\end{align}\]
We know the total energy in the simple harmonic motion is given by –
\[E=\dfrac{1}{2}m{{(A\omega )}^{2}}\]
It is given the total energy to be 100J.
\[\Rightarrow \dfrac{1}{2}m{{(A\omega )}^{2}}=100J\]
Now, also the kinetic energy of the particle when the particle is at a point x is –
\[KE=\dfrac{1}{2}m{{\omega }^{2}}({{A}^{2}}-{{x}^{2}})\]
The position of the particle as per the question is at 0.707A. It can be given as –
\[x=0.707A=\dfrac{A}{\sqrt{2}}\]

The kinetic energy can be given by –
\[\begin{align}
  & \Rightarrow KE=\dfrac{1}{2}m{{\omega }^{2}}({{A}^{2}}-{{(\dfrac{A}{\sqrt{2}})}^{2}}) \\
 & \Rightarrow KE=\dfrac{1}{2}m{{\omega }^{2}}(\dfrac{2{{A}^{2}}-{{A}^{2}}}{2}) \\
 & \Rightarrow KE=\dfrac{1}{2}m{{\omega }^{2}}\dfrac{{{A}^{2}}}{2} \\
\end{align}\]
We already know that total Kinetic energy is 100J. Therefore, we can find the kinetic energy at the position x is given as –
\[\begin{align}
  & KE=\dfrac{1}{2}m{{\omega }^{2}}\dfrac{{{A}^{2}}}{2} \\
 & \Rightarrow KE'=\dfrac{KE}{2}=\dfrac{100}{2}=50J \\
\end{align}\]
The kinetic energy of the particle undergoing simple harmonic motion at 0.707A is 50J.

So, the correct answer is “Option B”.

Note:
 The kinetic energy and the potential energy of a simple harmonic oscillates between a given maximum and zero. The total energy of the system will always be constant. In our situation, the Kinetic energy is half the total energy, i.e., the other half energy is the potential energy.