Correct option is (2) 2n
Equation of displacement of particle executing SHM is given by
x = A sin(ωt + ϕ) ......(1)
Potential energy of a particle executing SHM is given by
U = \(\frac12\)kx2 = \(\frac12\)kA2 sin2(ωt + ϕ)
U = \(\frac14\)kA2 - \(\frac14\)kA2 cos(2ωt + 2ϕ) ....(2)
From (1) and (2),
It is clear that the time period of x = A sin(ωt + ϕ) is
T1 = \(\frac {2\pi}\omega\)
⇒ frequency, n = \(\frac \omega{2\pi}\)
While time period of x2 = A2 sin2(ωt + ϕ) is
T2 = \(\frac{2\pi}{2\omega}\) = \(\frac{\pi}{\omega}\)
⇒ frequency, n′ = \(\frac{\omega}\pi\)
∴ n′ = 2n