Question

# The maximum velocity of a particle, executing simple harmonic motion with an amplitude $$7\ mm$$, is $$4.4\ m/s$$. The period of oscillation is

A
100 s
B
0.01 s
C
10 s
D
0.1 s

Solution

## The correct option is B $$0.01\ s$$For a SHM, $$x=A\sin(\omega t +\phi)$$Velocity, $$v=\dfrac{dx}{dt}=A\omega \cos(\omega t+\phi)$$So, $$v_{max}=A\omega$$or $$4.4=7\times 10^{-3}(2\pi/T)$$or $$T=9.99\times 10^{-3}=10\times 10^{-3}=0.01 s$$Physics

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