Question

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The position of a simple harmonic oscillator is given by y = A sin(2πft), with f = 253 Hz. Find a value of t at which the potential energy is one quarter of its maximum value.

Answer 1

The potential of a simple harmonic oscillator with force constant \(k\) is \[U = \frac12 k y^2\] The maximum potential energy is found by inserting the maximum value of \(y\). In this case it's \(y_\text{max} = A\)

so a quarter of the maximum potential energy is \[\frac14 \frac12 k A^2 = \frac18 k A^2\]

hence solve for \[y = \frac12 A\]