Question

A glider is oscillating in SHM on an air track with an amplitude $A$. You slow it so that its amplitude becomes half. Find the total mechanical energy in terms of the previous value.
     $A)\text{ }1/2$
     $B)\text{ }1/3$
     $C)\text{ }1/\sqrt{5}$
     $D)\text{ }1/4$

Answer 1

Hint: This problem can be solved by using the direct formula for the total mechanical energy for a simple harmonic motion (SHM) in terms of the amplitude, the mass of the body and the angular frequency.
Formula used:
$E=\dfrac{1}{2}m{{\omega }^{2}}{{A}^{2}}$

Complete answer:
We will solve this problem using the direct formula for the total mechanical energy of a body in simple harmonic motion. So, let us write the formula.
The total mechanical energy $E$ of a body of mass $m$ in SHM of amplitude $A$ and angular frequency $\omega $ is given by
$E=\dfrac{1}{2}m{{\omega }^{2}}{{A}^{2}}$ --(1)
Now, let us analyze the question.
Let the mass of the glider be $m$ and the angular frequency of its SHM be $\omega $.
The initial amplitude of motion is ${{A}_{i}}=A$.
Let the initial total mechanical energy be ${{E}_{i}}$.
Now the amplitude is made half. So, the final amplitude is ${{A}_{f}}=\dfrac{A}{2}$.
Let the final total mechanical energy be ${{E}_{f}}$.
Now, using (1), we get
${{E}_{i}}=\dfrac{1}{2}m{{\omega }^{2}}{{A}_{i}}^{2}=\dfrac{1}{2}m{{\omega }^{2}}{{A}^{2}}$ --(2)
${{E}_{f}}=\dfrac{1}{2}m{{\omega }^{2}}{{A}_{f}}^{2}=\dfrac{1}{2}m{{\omega }^{2}}{{\left( \dfrac{A}{2} \right)}^{2}}=\dfrac{1}{2}m{{\omega }^{2}}\dfrac{{{A}^{2}}}{4}$ --(3)
Now, using (2) and (3), we get
$\dfrac{{{E}_{f}}}{{{E}_{i}}}=\dfrac{\dfrac{1}{2}m{{\omega }^{2}}\dfrac{{{A}^{2}}}{4}}{\dfrac{1}{2}m{{\omega }^{2}}{{A}^{2}}}=\dfrac{\dfrac{1}{4}}{1}=\dfrac{1}{4}$
$\therefore {{E}_{f}}=\dfrac{1}{4}{{E}_{i}}$
Hence, in terms of the previous value of the total mechanical energy, the final value of the total mechanical energy is ${{\dfrac{1}{4}}^{th}}$ of the previous.

Therefore, the correct answer is $D)\text{ }1/4$.

Note:
This problem could also have been solved by realizing the fact that the total mechanical energy of an SHM is directly proportional to the square of its amplitude and therefore, the ratio of the final energy to the initial energy would have been nothing but the ratio of the square of the final amplitude to the square of the initial amplitude. This would have allowed us to do away with the unnecessary variables of mass and angular frequency (which anyway get cancelled in the calculations as they remain constants in this case) and made our calculations a bit cleaner and simpler.