A block of mass m compresses a spring of stiffness k through a distance $l / 2$ as shown in the figure. If the block is not fixed to the spring, the period of motion of the block is

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

$2 π \sqrt{m / k}$

$(π + 4) \sqrt{m / k}$

$(1 + π) \sqrt{m / k}$

None of these

answer is B.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

The period of oscillation = $2 π \sqrt{\frac{m}{k}}$
$∴$ The period of motion till the block is in contact with the spring is $t_{1} = π \sqrt{\frac{m}{k}}$
Then, it leaves the spring with a speed.
$v = ω A = (\sqrt{\frac{k}{m}}) (\frac{1}{2})$
Then, it moves with constant velocity v for a distance $D = l + l = 2 l$
The corresponding time of motion.
$t_{2} = \frac{2 l}{v} = \frac{2 l}{\frac{1}{2} \sqrt{\frac{k}{m}}} = 4 \sqrt{\frac{m}{k}}$
$∴$ Time period of motion.
$t = t_{1} + t_{2} = π \sqrt{\frac{m}{k}} + 4 \sqrt{\frac{m}{k}}$
Hence, the correct answer is option (b).