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A particle performing SHM has a maximum velocity 20cm/s and maximum acceleration 80cm/s2 . Its amplitude will be:
(A) 10cm
(B) 2cm
(C) 5cm
(D) 8cm

Answer
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Hint The maximum velocity of a particle in simple harmonic motion is given by v=rω and the maximum acceleration of a particle in simple harmonic motion is given by a=rω2 . Divide one by the other and find the value ω . Use this value in either of the equations to find the value of the amplitude of the particle in simple harmonic motion.

Complete Step by step answer
Let the maximum velocity of the particle in simple harmonic motion be v . Let the maximum acceleration of the particle be a . Let the amplitude of the particle be r . Let ω be the frequency of the particle executing simple harmonic motion. It is given in the question that the maximum velocity of the particle is 20cm/s and the maximum acceleration of the particle is 80cm/s2 .
The maximum velocity of a particle executing simple harmonic motion is given by
v=rω
Substituting the value of maximum velocity, we get
20=rω
The maximum acceleration of a particle executing simple harmonic motion is given by
a=rω2
Substituting the maximum value of acceleration, we get
80=rω2
Dividing the above equation for maximum acceleration by the already written equation for maximum velocity, we get
8020=ω
ω=4Hz
Substituting this value of frequency into the equation for maximum velocity, we get
20=r×4
By cross multiplication, we get
r=204
r=5cm
That is, the maximum amplitude calculated is 5cm .

Hence, option (C) is the correct option.

Note
Simple harmonic motion is a type of periodic motion. In simple harmonic motion, the amplitude of a particle will also be its maximum displacement from the mean position of the particle. The particle will have maximum velocity when it is at the mean position and the particle will have maximum acceleration when it is at its extreme positions.
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