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(B) The displacement of the particle is given by $x = A \sin(\omega t)$.At half the amplitude,the displacement is $x = A/2$.The potential energy $(U)$

Vedclass · Accepted Answer

$3$

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(B) The displacement of the particle is given by $x = A \sin(\omega t)$.At half the amplitude,the displacement is $x = A/2$.The potential energy $(U)$ of a particle in $S.H.M.$ is given by $U = \frac{1}{2} k x^2$.Substituting $x = A/2$,we get $U = \frac{1}{2} k (A/2)^2 = \frac{1}{8} k A^2$.The total energy $(E)$ of the particle is $E = \frac{1}{2} k A^2$.The kinetic energy $(K)$ is given by $K = E - U = \frac{1}{2} k A^2 - \frac{1}{8} k A^2 = \frac{3}{8} k A^2$.Now,the ratio of kinetic energy to potential energy is $K/U = (\frac{3}{8} k A^2) / (\frac{1}{8} k A^2) = 3/1$.Therefore,the ratio is $3: 1$.

$A$ particle is executing linear $S.H.M.$ starting from the mean position. The ratio of the kinetic energy to the potential energy of the particle at a point of half the amplitude is (in $: 1$)

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