Questions
An object of mass 0.2 kg executes simple harmonic along X-axis with frequency of . At the position x = 0.04 m, the object has kinetic energy of 0.5 J and potential energy of 0.4 J. The amplitude of oscillation in meter is equal to
detailed solution
Correct option is B
E=12mω2A2⇒E=12m(2πf)2A2⇒A=12πf2Em Putting E=K+U we obtain, A=12π25π2×(0.5+0.4)0.2⇒A=0.06mTalk to our academic expert!
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When the displacement is half the amplitude, the ratio of potential energy to the total energy is
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