Question

When the displacement in SHM is equal to 1/4 of the amplitude x_m, what fraction of the total energy is (a) kinetic energy and (b) potential energy? (c) At what displacement, in terms of the amplitude, is the energy of the system half kinetic energy and half potential energy? (Give the ratio of the answer to the amplitude)

Answer 1

Total energy in SHM is given by:

E_max = (1/2)*k*xm^2

xm = Amplitude of SHM

Potential energy in SHM is given by:

PE = (1/2)*k*x^2

where x = displacement

Now when displacement is equal to (1/4) of amplitude, then (1st we need to solve part B)

when x = (1/4)*xm, then

PE = (1/2)*k*(xm/4)^2 = k*xm^2/32

Now ratio of Potential energy and Total energy will be:

PE/E_max = (k*xm^2/32)/(k*xm^2/2) = 1/16

PE/E_max = 0.0625 [Ans of PART B]

Now Kinetic energy at x displacement will be:

KE = E_max - PE

Ratio of Kinetic energy and max energy will be:

KE/E_max = (E_max - PE)/E_max = 1 - PE/E_max

KE/E_max = 1 - 0.0625 = 0.9375

Part C.

We need amplitude when KE = PE

We also know that

KE + PE = E_max

Since KE = PE, So

2PE = E_max

2*(1/2)*k*x^2 = (1/2)*k*xm^2

x^2 = xm^2/2

x = xm/sqrt 2

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