Question
A particle undergoes SHM with angular frequency $\omega .$ The initial displacement is $x_{0}$ and the initial velocity is $v_{0} .$ Deduce that an expression for the amplitude of this motion is $$A=\sqrt{x_{0}^{2}+\frac{v_{0}^{2}}{\omega^{2}}}$$.
Step 1
Step 1: We start with the equations for position and velocity in simple harmonic motion: \[x = A \sin(\omega t)\] \[v = \omega A \cos(\omega t)\] Show more…
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Round 1
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