Question

A particle executes $SHM$ with time period $T$ and amplitude $A$, The maximum possible average velocity in time $T/4$ is

• A. $\frac{2A}{T}$
• B. $\frac{4A}{T}$
• C. $\frac{8A}{T}$
• D. $\frac{4\sqrt{2A}}{T}$

Answer 1

The correct option is D $\frac{4\sqrt{2A}}{T}$

b'

Average velocity is given by $V_{avg}=dfracst$

When $t=dfracT4$ ,$V_{avg}=dfracsdfracT4=dfrac4sT$

So, we have to maximise the displacement.

$s$ is the component of line $AB$ along $y-$axis.

To maximise $s$ we have to maximise the component of $AB$ along $y-$axis,

So line $AB$ will be parallel to $y-$axis

$2theta=dfracpi2$(from the figure)

$Rightarrowtheta=dfracpi4$

$S=Asintheta+Asintheta$

$=2Asindfracpi4$

$=dfrac2Asqrt2$

$=dfrac2Asqrt2timesdfracsqrt2sqrt2=sqrt2A$

$therefore{v}_{avg}=dfrac4sT=dfrac4sqrt2AT$

Hence $v_{avg}=dfrac4sqrt2AT$

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