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?

force is mass*acceleration.

we have a very small mass: 11.4*0.001 kg

if we want 0.035n of force, we need an acceleration:

0.035 = 11.4*0.001*a

a = 0.035/(11.4*0.001) m/s^2

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 [tex]\omega =2\pi f=2\times 3.14\times 1.65=10.362rad/sec[/tex]

Angular frequency is equal to [tex]\omega =\sqrt{\frac{k}{m}}[/tex]

[tex]10.362 =\sqrt{\frac{k}{0.041}}[/tex]

Squaring both side

[tex]107.371 ={\frac{k}{0.041}}[/tex]

k = 4.40 N/m

For vertical osculation

[tex]mg=kA[/tex]

[tex]0.041\times 9.8=4.40\times A[/tex]

A = 0.091 m

So amplitude will be equal to 0.0391 m

explanation: magnetic pole, region at each end of a magnet where the external magnetic field is strongest. the south-seeking pole, or any pole similar to it, is called a south magnetic pole. unlike poles of different magnets attract each other; like poles repel each other.