Question:
In an engine the piston undergoes vertical simple harmonic motion with amplitude 7 cm. A washer rests on top of the piston and moves with it. The motor speed is slowly increased. The frequency of the piston at which the washer no longer stays in contact with the piston, is close to :
In an engine the piston undergoes vertical simple harmonic motion with amplitude 7 cm. A washer rests on top of the piston and moves with it. The motor speed is slowly increased. The frequency of the piston at which the washer no longer stays in contact with the piston, is close to :
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Contact is lost when the downward acceleration of the piston at the top of its motion reaches g, i.e. \(\omega^2 A = g\).
Updated On: Nov 11, 2026
- \(0.1\ \mathrm{Hz}\)
- \(1.2\ \mathrm{Hz}\)
- \(0.7\ \mathrm{Hz}\)
- \(1.9\ \mathrm{Hz}\)
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The Correct Option is D
Solution and Explanation
The washer leaves the piston when the piston's downward acceleration exceeds g. Maximum acceleration is \(\omega^2 A\), reached at the top.
Condition: \(\omega^2 A = g\) so \(\omega = \sqrt{g/A} = \sqrt{9.8/0.07} = \sqrt{140} \approx 11.8\ \mathrm{rad/s}\).
\(f = \omega/2\pi = 11.8/6.28 \approx 1.88\ \mathrm{Hz} \approx 1.9\ \mathrm{Hz}\).
Answer: (4) \(1.9\ \mathrm{Hz}\)
Condition: \(\omega^2 A = g\) so \(\omega = \sqrt{g/A} = \sqrt{9.8/0.07} = \sqrt{140} \approx 11.8\ \mathrm{rad/s}\).
\(f = \omega/2\pi = 11.8/6.28 \approx 1.88\ \mathrm{Hz} \approx 1.9\ \mathrm{Hz}\).
Answer: (4) \(1.9\ \mathrm{Hz}\)
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