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Q.

A particle executes SHM with time period T and amplitude A. The maximum possible average velocity in time $\frac{T}{4}$ is:

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a

\frac{2 A}{T}

b

\frac{4 A}{T}

c

\frac{6 A}{T}

d

\frac{4 \sqrt{2} A}{T}

answer is D.

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Detailed Solution

Concept: To solve this question, we need to use the basic theory of simple harmonic motion (SHM). First suppose, SHM is represented by x=A sin wt and then we directly use some basic formula so that we will get the answer.
Formula used- The maximum possible average velocity =

\frac{maximum displacement}{time}

i.e. Vavg

here

avg

Hence, option 4 is correct.

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