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Question

The equation of a particle executing SHM is x=(5m)sin[(π s1)t+π3]. What is its amplitude, time period, maximum speed and velocity. At t = 1s?


A

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B

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C

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D

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Solution

The correct option is B


A typical equation representing SHM looks like x=A sin(ω t+ψ0)

Where x represents displacement

ω represents angular frequency

ψ0 represents initial phase

A represents amplitude

In the question we have x=(5m)sin[(π s1)t+π3]

So let's compare both! In a simple glance one can see

A=5m, ω=π s1 and ψ0=π3

Time period 2πω=2ππ=2s

Now what about velocities?

Let's differentiate the equation dxdt=(5m)π cos{(π s1)t+π3}

v=(5π)mcos{(π s1)t+π3}

We know that dxdt means velocity. To find the maximum velocity we can argue that cos has maximum value of 1 so Vmax=5π ms1. To

find velocity at t = 1s, let's substitute for t in velocity equation we get V1s=5π ms12.


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