Maximum potential energy of a block executing shm is 25J a is amplitude of oscillation At A/2 the kinetic energy of the block is

Question

The maximum potential energy of a block executing simple harmonic motion is 25 J.  A is amplitude of oscillation. At A/2, the kinetic energy of the block is

Answer 1

The maximum potential in an S.H.M is given by 

\(\frac{1}{2}kA^ 2=25\)

Potential energy of the simple harmonic oscillation at x is given by 

\(U =\frac{1}{2}kx^ 2\)

At x = A/2, U is

⇒ \(\frac{1}{2}K[\frac{A}{2}]^2=\frac{25}{4}\)

We know total kinetic energy is given by T = K + U. 

Also, T is equal to the maximum potential energy. 

∴ K + U = 25

Therefore, at U = 25/4

⇒ K = 25 - 25/4 = 18.75J