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A particle is performing SHM with amplitude A & time T then maximum Average velocity for time interval T/4

Question :- A particle is performing SHM with amplitude A & time T then maximum Average velocity for time interval T/4.

(A) $\displaystyle \frac{2\sqrt{2}A}{T}$

(B) $\displaystyle \frac{3\sqrt{2}A}{T}$

(D) $\displaystyle \frac{4\sqrt{2}A}{T}$

Solution :-

According to question we have to find “maximum” average velocity for the given time interval, i.e., T/4. So we have to consider those two points between which the particle moves “fastest” so that the average velocity comes out to be maximum. Those two points must be symmetrically placed around the mean position (x = 0) , because at mean position the velocity of the particle exhibiting SHM is maximum.

Let us split the given time interval T/4 into two equal parts T/8 and T/8 and find the position of the particle at time t = T/8 using equation of SHM :-

x = Asinωt

$\displaystyle \Rightarrow x=Asin\left( \omega \times \frac{T}{8} \right)=Asin\left( \frac{2\pi }{T}\times \frac{T}{8} \right)=Asin\left( \frac{\pi }{4} \right)$

$\displaystyle \therefore x=+\frac{A}{\sqrt{2}}$

This position is indicated by point “Q” in below gif :-

Similarly the point “P” indicates the position $\displaystyle x=-\frac{A}{\sqrt{2}}$ on the left hand side of mean position. In between these two points the particle take a total time of ( T/8) + (T/8) = T/4 and it moves fastest leading to maximum average velocity.

$\displaystyle Maximum\text{ }Average\text{ }Velocity\text{ }=~\frac{Total\text{ }Displacement}{Total\text{ }Time}$

$\displaystyle \Rightarrow Maximum\text{ }Average\text{ }Velocity\text{ }=\frac{\left[ \frac{A}{\sqrt{2}}+\frac{A}{\sqrt{2}} \right]}{\frac{T}{4}}=\frac{4\sqrt{2}A}{T}$

Hence (D) is the correct option.

A particle is performing SHM with amplitude A & time T then maximum Average velocity for time interval T/4

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