Question

Figure shown as block of mass m resting on a smooth horizontal ground attached to one end of a spring of force constant k in natural length. If another block of same mass and moving with a velocity u toward right is placed on the block which stick to it due to friction. The time it will take to reach its extreme position. Also find the amplitude of oscillations of the combined mass 2m

• A. $A=\sqrt{\frac{m}{2k}},t=\sqrt{\frac{k}{2m}}$
• B. $A=u\sqrt{\frac{m}{2k}},t=\frac{\pi }{2}\sqrt{\frac{2m}{k}}$
• C. $A=\sqrt{\frac{um}{2k}},t=\pi \sqrt{\frac{2m}{k}}$
• D. none of these

Answer 1

The correct option is C $A=\sqrt{\frac{um}{2k}},t=\pi \sqrt{\frac{2m}{k}}$

Given,

$\begin{array}{c}T=\frac{2\pi }{w}T=2\pi \sqrt{\frac{2m}{k}}\\ w=\sqrt{\frac{k}{m}},w^1=\sqrt{\frac{k}{2m}}\end{array}$

And the time taken to reach from mean position to extreme position is $\frac{f}{4}$.

So,$t=\frac{T^1}{4}=\frac{\pi }{2}\sqrt{\frac{2m}{k}}$

And, $V=A.\sqrt{\frac{k}{m}}\left\{\therefore w=\sqrt{\frac{k}{m}}\right\}$

$\begin{array}{c}\frac{V}{2}=A^1.\sqrt{\frac{k}{2m}}\\ \sqrt{2}=\frac{A\sqrt{2}}{A^{\prime }}\\ A^{\prime }=\frac{A}{\sqrt{2}}=\frac{\mu }{\sqrt{2}}\sqrt{\frac{m}{k}}=\mu \sqrt{\frac{m}{2k}}\end{array}$

$\therefore$ Option $C$ is correct.