Question

A block of mass $M$ is connected to one end of a spring of constant $k$. The other end is connected to the wall. Another block of mass $m$ is placed on $M$. The coefficient of static friction is $\mu$. Find the maximum amplitude of oscillation so that the block of mass $m$ does not slip on the lower block.

• A. $\frac{\mu mg}{k}$
• B. $\frac{\mu Mg}{k}$
• C. $\frac{\mu (M+m)g}{k}$
• D. $\mu \left(\frac{M}{M+m}\right)g/k$

Answer 1

The correct option is C $\frac{\mu (M+m)g}{k}$

Angular frequency of the system is

$\omega =\sqrt{\frac{k}{m+M}}$

The upper block of mass $m$ will not slip over the lower block of mass $M$ if the maximum force on the upper block $f_{max}$ does not exceed the frictional force $\mu mg$ between the two blocks.

So we have,

$f_{max}=m{a}_{max}=m{\omega }^2{A}_{max}\Rightarrow f_{max}=m{\left(\sqrt{\frac{k}{m+M}}\right)}^2{A}_{max}\Rightarrow f_{max}=\frac{mk{A}_{max}}{m+M}$

Hence, for no slipping,

$f_{max}=\mu mg\Rightarrow \frac{mk{A}_{max}}{m+M}=\mu mg\Rightarrow A_{max}=\frac{\mu (M+m)g}{k}$