Question

A body is executing simple harmonic motion with frequency 'n', the frequency of its potential energy is : -

(1) n

(2) 2n

(3) 3n

(4) 4n

Answer 1

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

Answer 2

Option : (2) 2n

Displacement equation of SHM of frequency 'n' x = Asin(ωt) = Asin(2πnt) 

Now,

So frequency of potential energy = 2n

Answer 3

Time period = T

Then frequency S.H.M. n = \(\frac1T\)

Frequency of potential energy n' = 1/T

\(\because\) T/2

n' = \(\frac1{T/2}\)

n'= 21/T

n' = 2n