Question

The coefficient of friction between the blocks of mass $m$ and $2m$ is $\mu =2\tan \theta$. There is no friction between the mass $2m$ and inclined plane. The maximum amplitude of the two block system for which there is no relative motion between both the blocks is:

• A. $\frac{mg\sin \theta }{K}$
• B. $\frac{3mg\sin \theta }{K}$
• C. $g\sin \theta \sqrt{\frac{K}{m}}$
• D. $\frac{2mg\sin \theta }{K}$

Answer 1

The correct option is B $\frac{3mg\sin \theta }{K}$

The maximum tendency of slipping is at the extreme point, as acceleration is maximum at the extreme position.
Maximum frictional force between $m$ and $2m$ is given by
$f_{\text{max}}=f_L=\mu N=\mu mg\cos \theta$
At extreme position, from FBD, we get,
$f-mg\sin \theta =m{\omega }^{2}A$
where ${\omega }^{2}A=a_{\text{max}}$


$\Rightarrow f=m{\omega }^{2}A+mg\sin \theta$
For zero relative motion between the blocks, we know that, $f\le f_{\text{max}}$
$\Rightarrow m{\omega }^{2}A+mg\sin \theta \le \mu mg\cos \theta$
From the data given in the question,
$m{\omega }^{2}A+mg\sin \theta \le \left(2\tan \theta \right)mg\cos \theta$$[\because \mu =2\tan \theta ]$
$\Rightarrow m{\omega }^{2}A\le mg\sin \theta$
$\Rightarrow A_{\text{max}}=\frac{g\sin \theta }{{\omega }^2}$
$\Rightarrow A_{\text{max}}=\frac{g\sin \theta }{\frac{K}{3m}}=\frac{3mg\sin \theta }{K}$
$[\because \omega =\sqrt{\frac{K}{3m}}]$
Thus, option (b) is the correct answer.