Gujarati
Hindi
13.Oscillations
medium

A mass at the end of a spring executes harmonic motion about an equilibrium position with an amplitude $A.$ Its speed as it passes through the equilibrium position is $V.$ If extended $2A$ and released, the speed of the mass passing through the equilibrium position will be

A

$2V$

B

$4V$

C

$\frac{V}{2}$

D

$\frac{V}{4}$

Solution

At equilibrium position the speed of a pendulum is equal to $A \omega$. If we double the amplitude, the speed will also double. so the correct option is $'A'$

Standard 11
Physics

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