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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 :

a. 0.7 Hz
b. 1.9 Hz
c. 1.2Hz
d. 0.1 Hz

1 Answer

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Final answer:

The frequency at which the washer will lose contact with the piston undergoing simple harmonic motion is 1.9 Hz. This is calculated by setting the maximum acceleration of the piston equal to gravitational acceleration and solving for the frequency.

Step-by-step explanation:

In the context of physics, when dealing with simple harmonic motion (SHM), there is a critical frequency beyond which the object on top of the piston will no longer maintain contact. This frequency corresponds to the point at which the acceleration of the piston exceeds the acceleration due to gravity. The acceleration in SHM is given by ω2X, where ω is the angular frequency and X is the amplitude. So, to find the frequency at which the washer will lose contact, we set the maximum acceleration equal to the acceleration due to gravity (g), which gives us ω2X = g, and since ω = 2πf, this equates to (2πf)2X = g. Solving for f (frequency) we get:

f = (1/2π)√(g/X)

Given that the amplitude (X) is 7 cm or 0.07 m and g = 9.81 m/s2, we can calculate the frequency as follows:

f = (1/2π)√(9.81/0.07)

f = (1/2π)√(140.14)

f = (1/2π)(11.84)

f ≈ 1.9 Hz

Therefore, the correct option is b. 1.9 Hz.

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