When the displacement in SHM Is equal to 1/2 of the amplitude XM, what fraction

Question

When the displacement in SHM Is equal to 1/2 of the amplitude XM, 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) When the displacement in SHM Is equal to 1/2 of the amplitude XM, 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)

Explanation / Answer

If Xm is the amplitude,

Total energy, TE = (1/2)*k*Xm^2

a) Kinetic energy, KE = Total energy - potentail energy

= (1/2)*k*Xm^2 - (1/2)*k*x^2

= (1/2)*k*Xm^2 - (1/2)*k*(Xm/2)^2

= 0.375*k*Xm^2

so, KE/TE = 0.375*k*Xm^2/(0.5*k*Xm^2)

= 0.75

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

= (1/2)*k*(Xm/2)^2

= 0.125*k*Xm^2

PE/TE = 0.125*k*Xm^2/(0.5*k*Xm^2)

= 0.25

c) Apply, (1/2)*k*x^2 = (1/2)*(1/2)*k*Xm^2

x = Xm/sqrt(2)

= 0.707*Xm

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