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Question

The displacement of a particle executing simple harmonic motion is given by
x=3sin[2πt+π4] where x is in metres and t is in seconds. The amplitude and maximum speed of the particle is:

A
3m,2πms1
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B
3m,4πms1
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C
3m,6πms1
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D
3m,8πms1
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Solution

The correct option is C 3m,6πms1
The given equation of SHM is

x=3sin[2πt+π4]

Compare the given equation with standard equation of SHM
x=Asin(ωt+ϕ)

we get, A=3m,ω=2πs1

Maximumspeed,vmax=Aω=3m×2πs1=6πms1

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