Question

A bob of mass $2m$ hangs by a string attached to the block of mass $m$ of a spring block system. The whole arrangement is in a state of equilibrium. The bob of mass $2m$ is pulled down slowly by a distance $x_0$ and released.Then:

• A. for $x_0=\frac{3mg}{k}$ , maximum tension in string is $4mg$
• B. for $x_0>\frac{3mg}{k}$ , maximum tension in string is $mg$
• C. frequency of oscillation of system is $\frac{1}{2\pi }\sqrt{\frac{k}{3m}}$ , for all non-zero values of $x_0$
• D. the motion will remain simple harmonic for $x_0\le \frac{3mg}{k}$

Answer 1

Answer: (A), (D)

The correct options are
A for $x_0=\frac{3mg}{k}$ , maximum tension in string is $4mg$
D the motion will remain simple harmonic for $x_0\le \frac{3mg}{k}$

While in equilibrium the string force in the string is

$kx-3mg=3ma$

And, $kx=3mg$

After the string is stretched and released the string will follow harmonic motion

For additional $x_o$ considering block and bob as one mass of 3m
Therefore after the release of the string

$k(x+x_o)-3mg=3ma$

$k{x}_o=3ma$

$a=\frac{k{x}_o}{3m}$

So the tension in the string is

$T-2mg=2ma$

$T=2mg+2m\times \frac{k{x}_o}{3m}$

for $x_o=\frac{3mg}{k}$

maximum tension is $T=4mg$

$a=g$ for $x_o=\frac{3mg}{k}$ which signifies the maximum limit of extension without any any external force applied.