Question

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

Answer 1

$\frac{3mg \sin \theta}{k}$

Answer 2

The system oscillates with angular frequency $\omega = \sqrt{\frac{k}{3m}}$. For block 'm', the forces along the incline are gravity ($mg \sin \theta$) and friction ($f_s$). The equation of motion for block 'm' is $ma = mg \sin \theta \pm f_s$. At the extreme positions, the acceleration is $a = \pm \omega^2 A$.

At the lower extreme (maximum upward acceleration): $m(\omega^2 A) = mg \sin \theta - f_s$ (friction acts downwards on 'm' to oppose upward motion) $f_s = mg \sin \theta - m\omega^2 A$

At the upper extreme (maximum downward acceleration): $m(-\omega^2 A) = mg \sin \theta + f_s$ (friction acts upwards on 'm' to oppose downward motion) $f_s = -mg \sin \theta - m\omega^2 A$

Let's re-evaluate the forces. Up the incline is positive. $a = -\omega^2 x$. At $x = -A$ (lower extreme), $a = \omega^2 A$ (upwards). Block 'm' tends to slide up relative to '2m'. So friction $f_s$ on 'm' is downwards. $ma = mg \sin \theta - f_s$ $m(\omega^2 A) = mg \sin \theta - f_s$ $f_s = mg \sin \theta - m\omega^2 A$

At $x = +A$ (upper extreme), $a = -\omega^2 A$ (downwards). Block 'm' tends to slide down relative to '2m'. So friction $f_s$ on 'm' is upwards. $ma = mg \sin \theta + f_s$ $m(-\omega^2 A) = mg \sin \theta + f_s$ $f_s = -mg \sin \theta - m\omega^2 A$

For no relative motion, $|f_s| \le f_{s,max}$. $f_{s,max} = \mu N' = (2 \tan \theta)(mg \cos \theta) = 2mg \sin \theta$.

From the lower extreme condition: $|mg \sin \theta - m\omega^2 A| \le 2mg \sin \theta$. $-2mg \sin \theta \le mg \sin \theta - m\omega^2 A \le 2mg \sin \theta$. The right inequality: $mg \sin \theta - m\omega^2 A \le 2mg \sin \theta \implies -m\omega^2 A \le mg \sin \theta$, which is always true. The left inequality: $-2mg \sin \theta \le mg \sin \theta - m\omega^2 A \implies m\omega^2 A \le 3mg \sin \theta \implies \omega^2 A \le 3g \sin \theta$.

From the upper extreme condition: $|-mg \sin \theta - m\omega^2 A| \le 2mg \sin \theta$. $|-(mg \sin \theta + m\omega^2 A)| \le 2mg \sin \theta$. $mg \sin \theta + m\omega^2 A \le 2mg \sin \theta$. $m\omega^2 A \le mg \sin \theta$. $\omega^2 A \le g \sin \theta$.

The more restrictive condition is $\omega^2 A \le g \sin \theta$. Maximum amplitude $A_{max}$ is when $\omega^2 A_{max} = g \sin \theta$. $A_{max} = \frac{g \sin \theta}{\omega^2} = \frac{g \sin \theta}{k/(3m)} = \frac{3mg \sin \theta}{k}$.