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Physics-

General

Easy

Question

A pendulum makes perfectly elastic collision with block of m lying on a frictionless surface attached to a spring of force constant k. Pendulum is slightly displaced and released. Time period of oscillation of the system is

$2 π [\sqrt{\frac{l}{g}} + \sqrt{\frac{m}{k}}]$
$π [\sqrt{\frac{l}{g}} + \sqrt{\frac{m}{k}}]$
$2 π \sqrt{\frac{l}{g}}$
$2 π \sqrt{\frac{m}{k}}$

The correct answer is: $pi open square brackets square root of fraction numerator l over denominator g end fraction end root plus square root of fraction numerator m over denominator k end fraction end root close square brackets$

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