Physics, 26.06.2020 15:01 kayolaaaa53

A block of mass M is suspended from a ceiling by a spring with spring stiffness constant k. A penny of mass m is placed on top of the block. What is the maximum amplitude of oscillations that will allow the penny to just stay on top of the block

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Answer from: princesskoi8561

Explanation:

Given that:

Mass of block M

Mass of penny m

spring stiffness constant k

The frequency of oscillation of the block $f=\frac{1}{2\pi}\sqrt{\frac{k}{m} }$

The angular velocity is $\omega =2\pi f$

$=\sqrt{k/m}$

when the penny is resting on the block

The acceleration of the penny = acceleration of the block

If R is the reaction of the block on the penny

$R-mg=a_{max}m\\\\=-\omega^2A_{max}m\\\\R=mg-\omega^2A_{max}m$

The penny will leave the block if R = 0

$mg=\omega^2A_{max}m\\\\g=\omega^2A_{max}\\\\A_{max}=\frac{g}{\omega^2} \\\\=\frac{g}{(k/M)} \\\\A_{max}=gM/k$

Therefore the amplitude $A_{max}$ for the penny to remain on the block

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