Question

A block of mass m rigidly attached with a spring k is compressed through a distance A. If the block is released, the period of oscillation of the block for a complete cycle is equal to

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

The period of motion from A to O is equal to quarter of the time period T of oscillation of mass spring system.
$\Rightarrow t_{AO}=\frac{T}{4}=\frac{1}{4}\left[2\pi \sqrt{\frac{m}{k}}\right]=\frac{\pi }{2}\sqrt{\frac{m}{k}}$
Since the motion is simple harmonics
$OB=OA\sin \frac{2\pi }{T}{t}_{OB},where{t}_{OB}$ is the time of motion from O to B.
$\Rightarrow t_{oB}=\frac{T}{2\pi }si{n}^{-1}\frac{\frac{A}{2}}{A}=\frac{T}{2\pi }\left(\frac{\pi }{6}\right)=\frac{T}{12}=\frac{2\pi }{12}\sqrt{\frac{m}{k}}=\frac{\pi }{6}\sqrt{\frac{m}{k}}$
$\therefore$ The total time of motion for a complete cycle = $t=2(t_{AQ}+t_{QB})$
$\Rightarrow t=2[\frac{\pi }{2}\sqrt{\frac{m}{k}}+\frac{\pi }{6}\sqrt{\frac{m}{k}}]=\frac{4\pi }{3}\sqrt{\frac{m}{k}}$