A body is executing simple harmonic motion with frequency ‘n’, the frequency of its potential energy is :
Answer
Correct Answer : b ) 2n
Explanation :A particle oscillating back and forth about the origin of an x-axis between certain limits +A and –A. This oscillatory motion is said to be simple harmonic if the displacement x of the particle from the origin varies with time as:
x (t) = A cos (ω t)
where A and ω are amplitude and angular frequency respectively.
The potential energy of a particle executing simple harmonic motion (SHM) is given by-
\(U(x)=m ω^2x^2\)
\(U(t)=m ω^2A^2 cos^2(ωt)\)
\(U(t)\)=\({1\over2}m ω^2A^2 {({{1+cos^2ωt}\over 2})}\)
ωPE= 2ω
ω is the angular frequency of SHM, x is the displacement as a function of time. If the particle is executing SHM with frequency n, then the frequency of its potential energy will be 2n.
Hence, option (B) is correct.
This question was asked in Physics Section of NEET Exam 2021.
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