Question

A mass \(0.4\) kg performs S.H.M. with a frequency \(\frac{16}{π}\) Hz. At a certain displacement it has kinetic energy \(2\) J and potential energy \(1.2\) J. The amplitude of oscillation is

• A. \(0.125\) m
• B. \(0.1\) m
• C. \(0.05\) m
• D. \(0.15\) m

Hint:

Total energy = KE + PE = (1/2) m omega^2 A^2 with omega = 2 pi f.

Answer 1

The Correct Option is A

Step 1: Use spring constant:
$k=m\omega^2=0.4\times1024=409.6$ N/m.

Step 2: Use total energy:
$E=\dfrac12kA^2$, so $A=\sqrt{\dfrac{2E}{k}}=\sqrt{\dfrac{6.4}{409.6}}$.

Step 3: Compute:
$\dfrac{6.4}{409.6}=0.015625$, so $A=0.125$ m. Option A.

Final Answer:
Total energy 3.2 J with omega = 32 rad/s gives A = 0.125 m. \[ \boxed{\text{(A) }0.125\ \text{m}} \]