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 $=\frac{1}{2}m{\omega }^2(a^2-y^2)$
$=E-\frac{1}{2}m{\omega }^2{y}^2$
$wheny=\frac{a}{2}$
$\Rightarrow K=E-\frac{1}{2}m{\omega }^2\left(\frac{a^2}{4}\right)=E-\frac{E}{4}$
$E=\frac{3E}{4}$