A particle starts oscillating simple harmonically from its mean position with time period ' T '. At time t = 12 T , the ratio of the potential energy to kinetic energy of the particle is [

Question

A particle starts oscillating simple harmonically from its mean position with time period ' T '. At time t = 12 T ​ , the ratio of the potential energy to kinetic energy of the particle is [ sin 3 0 ∘ = cos 6 0 ∘ = 0.5 , cos 3 0 ∘ = sin 6 0 ∘ = 2 3 ​ ​ ]

TestHub · Accepted Answer

The equation of displacement for a particle executing simple harmonic motion starting from the mean position is given by:

$x = A sin (ω t)$

Given $t = \frac{T}{12}$ and $ω = \frac{2 π}{T}$ , the displacement is:

$x = A sin (\frac{2 π}{T} \times \frac{T}{12}) = A sin (\frac{π}{6}) = A sin (3 0^{\circ}) = \frac{A}{2}$

The potential energy ( $U$ ) of the particle is:

$U = \frac{1}{2} k x^{2} = \frac{1}{2} k (\frac{A}{2})^{2} = \frac{1}{8} k A^{2}$

The total energy ( $E$ ) of the particle is:

$E = \frac{1}{2} k A^{2}$

The kinetic energy ( $K$ ) is the difference between total energy and potential energy:

$K = E - U = \frac{1}{2} k A^{2} - \frac{1}{8} k A^{2} = \frac{3}{8} k A^{2}$

The ratio of potential energy to kinetic energy is:

$\frac{U}{K} = \frac{\frac{1}{8} k A ^{2}}{\frac{3}{8} k A ^{2}} = \frac{1}{3}$

Thus, the required ratio is $1 : 3$ .

Answer: $1 : 3$

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