The displacement of a particle, executing simple harmonic motion with time period T , is expressed as x(t)=A ω t , where A is the amplitude. The maximum value of potential energy of this oscillator is found at t=T / 2 β

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For SHM with x(t) = A (ω t) , the potential energy is U = (1)/(2)mω^2 x^2 = (1)/(2)mω^2 A^2 ^2(ω t) . Maximum potential energy occurs when ^2(ω t) = 1 , i.e., when ω t = (π)/(2), (3π)/(2), ... The first maximum occurs at ω t = (π)/(2) , giving t = (π)/(2ω) = (π)/(2) · (T)/(2π) = (T)/(4) . Since this is given as t = (T)/(2β) . We have (T)/(4) = (T)/(2β) , so β = 2 .

The displacement of a particle, executing simple harmonic motion...

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