# Simple harmonic motion

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Question

# Two simple harmonic motions are given by ${\mathrm{x}}_{1}=\frac{\mathrm{a}}{2}\mathrm{sin}\mathrm{\omega t}+\frac{\sqrt{3}\mathrm{a}}{2}\mathrm{cos}\mathrm{\omega t}$ and ${\mathrm{x}}_{2}=\mathrm{asin}\mathrm{\omega t}+\mathrm{acos}\mathrm{\omega t}$ above, the minimum phase difference between the two simple harmonic motions is (in degrees)

Moderate
Solution

## Amplitude of the first simple harmonic motion is${\mathrm{A}}_{1}=\sqrt{{\left(\frac{\mathrm{a}}{2}\right)}^{2}+{\left(\frac{\sqrt{3}\mathrm{a}}{2}\right)}^{2}}=\mathrm{a}$Amplitude of the second motion isWe can rewrite the two motions asFrom (1) and (2) it follows that the amplitudes are a and $\frac{a}{\sqrt{2}}$ (a result obtained above) the phases are 60 and 45°. Therefore, phase difference = 60°- 45° = 15°

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