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 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
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Total energy = KE + PE = (1/2) m omega^2 A^2 with omega = 2 pi f.
Updated On: Oct 1, 2026
- \(0.125\) m
- \(0.1\) m
- \(0.05\) m
- \(0.15\) m
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Concept:
In SHM the total energy is constant: \(E=K+U=\dfrac12m\omega^2A^2\).
Step 2: Find the total energy and omega:
\(E=2+1.2=3.2\) J. \(\omega=2\pi f=2\pi\times\dfrac{16}{\pi}=32\) rad/s.
Step 3: Solve for the amplitude:
\(3.2=\dfrac12(0.4)(32)^2A^2=0.2\times1024\,A^2=204.8A^2\). So \(A^2=\dfrac{3.2}{204.8}=0.015625\) and \(A=0.125\) m. Option A.
Step 4: Why the other options are wrong.
0.1, 0.05 and 0.15 m would give total energies 2.05 J, 0.51 J and 4.6 J, none equal to 3.2 J.
Final Answer:
The amplitude is 0.125 m. \[ \boxed{\text{(A) }0.125\ \text{m}} \]
In SHM the total energy is constant: \(E=K+U=\dfrac12m\omega^2A^2\).
Step 2: Find the total energy and omega:
\(E=2+1.2=3.2\) J. \(\omega=2\pi f=2\pi\times\dfrac{16}{\pi}=32\) rad/s.
Step 3: Solve for the amplitude:
\(3.2=\dfrac12(0.4)(32)^2A^2=0.2\times1024\,A^2=204.8A^2\). So \(A^2=\dfrac{3.2}{204.8}=0.015625\) and \(A=0.125\) m. Option A.
Step 4: Why the other options are wrong.
0.1, 0.05 and 0.15 m would give total energies 2.05 J, 0.51 J and 4.6 J, none equal to 3.2 J.
Final Answer:
The amplitude is 0.125 m. \[ \boxed{\text{(A) }0.125\ \text{m}} \]
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