Simple Harmonic Motion

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A particle executes SHM with an amplitude of 5 cm. When the particle is at 4 cm from the mean position, the magnitude of its velocity is equal to that of its acceleration. What is the time period in seconds?

RankGuru · Accepted Answer

1. **Velocity in SHM**: $v = \omega \sqrt{A^2 - x^2}$.
2. **Acceleration in SHM**: $a = \omega^2 x$.
3. **Given condition**: $v = a$ at $x=4	ext{ cm}$ and $A=5	ext{ cm}$.
   $$\omega \sqrt{5^2 - 4^2} = \omega^2 (4)$$
   $$\sqrt{25 - 16} = 4\omega \implies 3 = 4\omega \implies \omega = \frac{3}{4}$$
4. **Time Period ($T$)**: $T = \frac{2\pi}{\omega} = \frac{2\pi}{3/4} = \frac{8\pi}{3}$.

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