Question

The total energy of the body executing S.H.M. is E. Then the kinetic energy when the displacement is half of the amplitude, is

[RPMT 1994, 96; CBSE PMT 1995; JIPMER 2002]

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

Total energy in SHM E = $\frac{1}{2}m{\omega }^2{a}^2$ : ( where a = amplitude )

Potential energy

U = $\frac{1}{2}m{\omega }^2(a^2-y^2)=E-\frac{1}{2}m{\omega }^2{y}^2$

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