Question

The friction coefficient between the two blocks shown in the figure is $\mu$ and the horizontal plane is smooth. Find the magnitude of the frictional force between the blocks when the displacement from the mean position is $x$.

• A. $\left(\frac{mk|x|}{M+\mu m}\right)$
• B. $\left(\frac{mk|x|}{\mu M+m}\right)$
• C. $\left(\frac{mk|x|}{M+m}\right)$
• D. $\left(\frac{\mu mk|x|}{M+m}\right)$

Answer 1

The correct option is C $\left(\frac{mk|x|}{M+m}\right)$

$\text{Given:-}$The friction coefficient between the two blocks $=\mu$

$\text{To find:-}$The magnitude of the frictional force between the blocks when the displacement from the mean position is x

$\text{Solution:-}$

For small amplitude, the two blocks oscillate together. The angular frequency is of a single block spring system:

$\omega =\sqrt{\left[\frac{k}{M+m}\right]}$

and so the time period is:

$T=2\pi \sqrt{\frac{M+m}{k}}$

When the blocks are at distance x from the mean position, their acceleration is:

$a=-{\omega }^{2}x=-\frac{kx}{M+m}$

The resultant force on the upper block is, therefore

$mf=-\frac{mkx}{M+m}$

The force is prompted by the frictional of the lower block. Hence the magnitude of the frictional force is

$\frac{mk\left|x\right|}{M+m}$

$\text{Hence the correct option is C}$