Question:
A mass \(m=1.0\) kg is put on a flat pan attached to a vertical spring fixed on the ground. The mass of the spring and the pan is negligible. When pressed slightly and released, the mass executes simple harmonic motion. The spring constant is 500 N/m. What is the amplitude A of the motion, so that the mass \(m\) tends to get detached from the pan ? (Take \(g=10\ \mathrm{m/s^2}\) ). The spring is stiff enough so that it does not get distorted during the motion. 
A mass \(m=1.0\) kg is put on a flat pan attached to a vertical spring fixed on the ground. The mass of the spring and the pan is negligible. When pressed slightly and released, the mass executes simple harmonic motion. The spring constant is 500 N/m. What is the amplitude A of the motion, so that the mass \(m\) tends to get detached from the pan ? (Take \(g=10\ \mathrm{m/s^2}\) ). The spring is stiff enough so that it does not get distorted during the motion.
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The mass tends to leave the pan at the highest point, where the pan's push becomes zero and the mass's downward acceleration equals \(g\): \(\omega^2A=g\) with \(\omega^2=k/m\).
Updated On: Oct 31, 2026
- \(A \gt 2.0\) cm
- \(A=2.0\) cm
- \(A \lt 2.0\) cm
- \(A=1.5\) cm
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The Correct Option is B
Solution and Explanation
At the top of the motion the acceleration is directed downward with magnitude \(\omega^2A\). For the mass: \(mg-N=m\omega^2A\).
The mass is just about to leave the pan when \(N=0\): \(\omega^2A=g\).
\(\omega^2=\frac km=\frac{500}{1.0}=500\ \mathrm{s^{-2}}\), so \(A=\frac{g}{\omega^2}=\frac{10}{500}=0.02\) m \(=2.0\) cm.
(For \(A \gt 2.0\) cm the mass would actually lose contact; at \(A=2.0\) cm it just tends to detach.)
Answer: (2) \(A=2.0\) cm
The mass is just about to leave the pan when \(N=0\): \(\omega^2A=g\).
\(\omega^2=\frac km=\frac{500}{1.0}=500\ \mathrm{s^{-2}}\), so \(A=\frac{g}{\omega^2}=\frac{10}{500}=0.02\) m \(=2.0\) cm.
(For \(A \gt 2.0\) cm the mass would actually lose contact; at \(A=2.0\) cm it just tends to detach.)
Answer: (2) \(A=2.0\) cm
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