A particle starts oscillating simple harmonically from its equilibrium position with time period 'T'. What is the ratio of potential energy to kinetic energy of the particle at time t = 12 T

Question

A particle starts oscillating simple harmonically from its equilibrium position with time period 'T'. What is the ratio of potential energy to kinetic energy of the particle at time t = 12 T ​ ? ( sin ( 6 π ​ ) = 2 1 ​ )

TestHub · Accepted Answer

The equation of displacement for a particle starting from its equilibrium position is $x = A sin (ω t)$ .

Given $t = \frac{T}{12}$ and $ω = \frac{2 π}{T}$ .

The phase angle is $ω t = \frac{2 π}{T} \times \frac{T}{12} = \frac{π}{6}$ .

The potential energy of the particle is $U = \frac{1}{2} m ω^{2} A^{2} sin^{2} (ω t)$ .

The kinetic energy of the particle is $K = \frac{1}{2} m ω^{2} A^{2} cos^{2} (ω t)$ .

The ratio of potential energy to kinetic energy is $\frac{U}{K} = tan^{2} (ω t)$ .

Substituting $ω t = \frac{π}{6}$ , we get $\frac{U}{K} = tan^{2} (\frac{π}{6}) = (\frac{1}{3})^{2} = \frac{1}{3}$ .

The ratio is $1 : 3$ .

Answer: $1 : 3$

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