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

Amplitude will be equal to 0.091 m

Step-by-step explanation:

Given mass of the slits = 41 gram = 0.041 kg

Frequency f = 1.65 Hz

So angular frequency
`\omega =2\pi f=2* 3.14* 1.65=10.362rad/sec`

Angular frequency is equal to
`\omega =\sqrt{(k)/(m)}`

`10.362 =\sqrt{(k)/(0.041)}`

Squaring both side

`107.371 ={(k)/(0.041)}`

k = 4.40 N/m

For vertical osculation

`mg=kA`

`0.041* 9.8=4.40* A`

A = 0.091 m

So amplitude will be equal to 0.0391 m