Question

# A particle performing simple harmonic motion has a frequency of $$1.2\ Hz$$ and a maximum acceleration of $$4\ m/s^2$$. Find The maximum velocity.

Solution

## Now, $$f = 1.2 \ Hz$$Now, Angular frequency $$\omega = 2\pi f = 2\pi \times 1.2 = 2.4\pi$$If A is the amplitude then maximum acceleration, $$a_{max} = A\omega ^2 = 4$$Then maximam velocity $$v_{max} = A\omega = \dfrac{a_{max}}{\omega} = \dfrac{4}{2.4\pi}=0.53 \ m/s^2$$Physics

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