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Question

A particle ís executing simple harmonic motion with amplitude of 0.1 m. At a certain instant when its displacement is 0.02 m, its acceleration is 0.5 ms2. The maximum velocity of the particle is (in ms1)


A
0.01
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B
0.05
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C
0.5
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D
0.25
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Solution

The correct option is C 0.5
Acceleration, a=0.5 ms2
Amplitude, A=0.1 m
Displacement, y=0.02 m
Using the formula of maximum acceleration a=ω2y
or 0.5=ω2×0.02 
or ω2=0.50.02=25
So, ω=5
Now, maximum velocity is
v=Aω=0.1×5=0.5 ms1

Physics

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