Question

The total energy of the body executing simple harmonic motion (SHM) is E. Then the kinetic energy when the displacement is half of the amplitude is

• A. $\frac{E}{2}$
• B. $\frac{E}{4}$
• C. $\frac{3E}{4}$
• D. $\frac{\sqrt 3 E}{4}$

Answer 1

Answer: (C)

TotaI energy in SHM
$e =\frac{1}{2}m\omega^2 a^2 ,(where , a=amplitude)$
Kinetic energy K $=\frac{1}{2} m \omega^2 (a^2 -y^2)$
$ \, \, \, \, \, \, \, \, \, \, \, \, =E -\frac{1}{2}m\omega^2 y^2$
$when \, y=\frac{a}{2} $
$\Rightarrow \, \, \, \, \, \, \, \, K=E -\frac{1}{2}m \omega^2 \bigg(\frac{a^2}{4}\bigg)=E-\frac{E}{4}$
$ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, E =\frac{3E}{4}$