Question

A small bolt with a mass of 41.0 g sits on top of a piston. The piston is undergoing simple harmonic motion in the vertical direction with a frequency of 1.65 Hz. What is the maximum amplitude with which the piston can oscillate without the bolt losing contact with the piston's surface?

Answer 1

Answer:

Amplitude will be equal to 0.091 m

Explanation:

Given mass of the slits = 41 gram = 0.041 kg

Frequency f = 1.65 Hz

So angular frequency $\omega =2\pi f=2\times 3.14\times 1.65=10.362rad/sec$

Angular frequency is equal to $\omega =\sqrt{\frac{k}{m}}$

$10.362 =\sqrt{\frac{k}{0.041}}$

Squaring both side

$107.371 ={\frac{k}{0.041}}$

k = 4.40 N/m

For vertical osculation

$mg=kA$

$0.041\times 9.8=4.40\times A$

A = 0.091 m

So amplitude will be equal to 0.0391 m