- Home
- Standard 11
- Physics
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
Similar Questions
hard