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Physics Question on Kinetic Energy

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

E2\frac{E}{2}

B

E4\frac{E}{4}

C

3E4\frac{3E}{4}

D

3E4\frac{\sqrt 3 E}{4}

Answer

3E4\frac{3E}{4}

Explanation

Solution

TotaI energy in SHM e=12mω2a2,(where,a=amplitude)e =\frac{1}{2}m\omega^2 a^2 ,(where , a=amplitude) Kinetic energy K =12mω2(a2−y2)=\frac{1}{2} m \omega^2 (a^2 -y^2)             =E−12mω2y2 \, \, \, \, \, \, \, \, \, \, \, \, =E -\frac{1}{2}m\omega^2 y^2 when y=a2when \, y=\frac{a}{2} ⇒        K=E−12mω2(a24)=E−E4\Rightarrow \, \, \, \, \, \, \, \, K=E -\frac{1}{2}m \omega^2 \bigg(\frac{a^2}{4}\bigg)=E-\frac{E}{4}                E=3E4 \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, E =\frac{3E}{4}