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. $\frac{5\pi }{3}\sqrt{\frac{m}{k}}$

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.

$\Rightarrow t_{AO}=\frac{T}{4}=\frac{1}{4}[2\pi \sqrt{\frac{m}{k}}]=\frac{\pi }{2}\sqrt{\frac{m}{k}}$

Since the motion is smple harmonic,

$OB=OAsin\frac{2\pi }{T}{t}_{OB}$, where $t_{OB}$ is the times of motion from O to B.

$\Rightarrow t_{OB}=\frac{T}{2\pi }si{n}^{-1}\frac{\frac{A}{2}}{A}=\frac{T}{2\pi }(\frac{\pi }{6})=\frac{T}{12}=\frac{2\pi }{12}\sqrt{\frac{m}{k}}=\frac{\pi }{6}\sqrt{\frac{m}{k}}$

$\therefore$ The total time of motion for a complete cycle $=t=2(t_{AO}+t_{OB})$

$\Rightarrow t=2[\frac{\pi }{2}\sqrt{\frac{m}{k}}+\frac{\pi }{6}\sqrt{\frac{m}{k}}]=\frac{4\pi }{3}\sqrt{\frac{m}{k}}$