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PhysicsSHMMiscellaneousMedium2 minPYQ_2020

PhysicsMediumsingle choice

A particle performs S.H.M. from the mean position. Its amplitude is 'A' and total energy is 'E'. At a perticular instant its kinetic energy is $\frac{3 E}{4}$ . The displacement of the particle at that instant is

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Solution:

Total energy $= E = \frac{1}{2} m ω^{2} A^{2}$ When kinetic energy is $\frac{3 E}{4}$ , its potential energy is $(E - \frac{3 E}{4}) = \frac{E}{4}$ Dividing Eq.(1) by Eq.(2), $4 = \frac{A ^{2}}{x ^{2}} ∴ x = \frac{A}{2}$

Stream:MHT_CETSubject:PhysicsTopic:SHMSubtopic:Miscellaneous

⏱ 2mℹ️ Source: PYQ_2020

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