Answer:
Explanation:
Given that:
Mass of block M
Mass of penny m
spring stiffness constant k
The frequency of oscillation of the block [tex]f=\frac{1}{2\pi}\sqrt{\frac{k}{m} }[/tex]
The angular velocity is [tex]\omega =2\pi f[/tex]
[tex]=\sqrt{k/m}[/tex]
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
[tex]R-mg=a_{max}m\\\\=-\omega^2A_{max}m\\\\R=mg-\omega^2A_{max}m[/tex]
The penny will leave the block if R = 0
[tex]mg=\omega^2A_{max}m\\\\g=\omega^2A_{max}\\\\A_{max}=\frac{g}{\omega^2} \\\\=\frac{g}{(k/M)} \\\\A_{max}=gM/k[/tex]
Therefore the amplitude [tex]A_{max}<gM/k[/tex] for the penny to remain on the block
