Question

A block of mass $M$ is attached with the springs as shown .If the block is slightly displaced the time period of $SHM$ for the block shown in the figure will be

• A. $2\pi \sqrt{\frac{m}{9k}}$
• B. $2\pi \sqrt{\frac{m}{k}}$
• C. $\frac{4\pi }{3}\sqrt{\frac{m}{9k}}$
• D. $\pi \sqrt{\frac{m}{2k}}$

Answer 1

Answer: (A)

When system moves x units in right then Pulley 1 and Pulley 2 moves xunits in right and required string for each pulley is 2x.

Therefore , both spring are elongated by 2x.

Force exerted$=k\left(2x\right)$

Tension in String(T)$=k\left(2x\right)$

Force exerted by right spring due to x displacement $=kx$

Net force $=2T+2T+kx=9kx$

Now net force $=m{\omega }^{2}x=9kx$

Or , $\omega =\sqrt{\frac{9k}{m}}$

Now time period for SHM$=T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{m}{9k}}$.