A particle of mass 2kg executing simple harmonic motion is 9J at a distance 0-8m from mean position. If the amplitude of vibration of particle is 1 m, then find the frequency of oscillation of particle.

Solution

We know that the kinetic energy of particle at a distance y from the mean position is,

space space space space space space space space space space space space space space space space space straight K equals 1 half mω squared left parenthesis straight A squared minus straight y squared right parenthesis

Given that,
Mass, m=2kg
Energy, K.E =9J
Amplitude, A = 1m
Dispalcement, y = 0.8m

therefore space space space space space 9 equals 1 half cross times 2 cross times straight omega squared open square brackets open parentheses 1 close parentheses squared minus left parenthesis 0.8 right parenthesis squared close square brackets space space space space space space space space space space space equals 0.36 space straight omega squared space rightwards double arrow space space space space space straight omega equals 5 space r a d divided by straight s space rightwards double arrow space space space space space straight v space equals space left parenthesis 5 divided by 2 straight pi right parenthesis space h z

This is the frequency of the oscillation of the particle.

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