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Question

# A particle executes SHM with an angular frequency 4π rad/s. If it is at its positive extreme position initially, find the first instant when it is at a distance of 1√2 times its amplitude from the mean position in positive direction

A
124 s
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B
116 s
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C
116 s
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D
316 s
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Solution

## The correct option is B 116 sLet A be the amplitude and ϕ be the phase constant for the oscillation. From the data given in the question, the displacment equation can be written as x=A sin (4πt+ϕ) ...(1) (∵ω=4π rad/s) Since at t=0;x=+A ∴4π(0)+ϕ=π2 ⇒ϕ=π2 ∴ equation (1) can be written as x=A sin (4πt+π2)=A cos(4πt) Evidently, A√2=A cos(4πt) ⇒4πt=π4 ⇒t=116 s

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