Question

The total energy of the body executing SHM is E. Then the kinetic energy when the displacement is half of the amplitude is?

Answer 1

Step 1: Given parameters:

The total energy of the body executing SHM = $E$

Step 2: To find:

The kinetic energy when the displacement is half of the amplitude KE=?

Step 3: Solution:

The total energy in SHM $E=\frac{1}{2}m{\omega }^2{a}^2$

The K.E of a body in SHM at displacement y from mean is $P.E=\frac{1}{2}m{\omega }^2\left(A^2-y^2\right)$

Displacement at half of the amplitude is $\frac{A}{2}$

The Kinetic energy when the displacement is half of the amplitude,

$KE=\frac{1}{2}m{\omega }^2\left(a^2-{\left(\frac{a}{2}\right)}^2\right)$

$KE=\frac{1}{2}m{\omega }^2\left(\frac{3}{4}{a}^2\right)$

$KE=\frac{3}{4}\left(\frac{1}{2}m{\omega }^2{a}^2\right)$

$KE=\frac{3}{4}E$

Hence, the kinetic energy when the displacement is half of the amplitude is $\frac{3}{4}E$.