First slide

Oscillation motion

Question

The total energy of a body executing SHM is E. Then the kinetic energy when its distance from equilibrium position is half of the amplitude, is

Moderate

A

E/2
B

E/4
C

3E/4
D

$\frac{\sqrt{3} E}{4}$

Solution

$\begin{array}{l} E = \frac{1}{2} {mA}^{2} ω^{2} \\ Now, KE = \frac{1}{2} {mA}^{2} ω^{2} (1 - \frac{x^{2}}{A^{2}}) \\ When x = \frac{A}{2}, KE = E [1 - \frac{{(A / 2)}^{2}}{A^{2}}] = \frac{3 E}{4} \end{array}$

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