Question

a body executes SHM with an amplitude xo its energy is half kinetic energy and half potential energy when displacement is.

Answer 1

The total energy of such body always remains constant, let this be EE.

E=K+UE=K+U, where K and UK and U are the kinetic and potential energies respectively.

If the total energy is half Kinetic and half potential, it is true only if K=UK=U.

Let the displacement of the body be xx.

The velocity of a body in SHM is given by:

v=ω(A2−x2−−−−−−√)v=ω(A2−x2), where,

ω=km−−−√ω=km, and kk is the force constant for the body and AA is the amplitude of the body.

The kinetic energy of the body is:

K=12mv2K=12mv2

 =12m(ω2(A2−x2)) =12m(ω2(A2−x2))

 =12mkm(A2−x2) =12mkm(A2−x2)

K=k2(A2−x2)…(i)K=k2(A2−x2)…(i)

The force done in increasing displacement of a body by x=F=−kxx=F=−kx

Let the increase in potential energy for changing the displacement by dxdx be −dU−dU.

Here, the negative sign indicates that the change occurs in a direction opposite to that of the change in the displacement.

Since force applied is conservative,

−dU=F.dx−dU=F.dx

⟹−dU=−k xdx⟹−dU=−k xdx

Integrating both sides,

∫U0dU=k∫x0xdx∫0UdU=k∫0xxdx

⟹U=12kx2⟹U=12kx2

Now, from the original condition, K=UK=U

Or,

12k(A2−x2)=12kx212k(A2−x2)=12kx2

⟹A2−x2=x2⟹A2−x2=x2

⟹2x2=A2⟹2x2=A2

⟹x2=A22⟹x2=A22

⟹x=±A2–√⟹x=±A2

At displacements ±A2–√±A2, the kinetic energy is equal to the potential energy.