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Q.

Maximum velocity in SHM is v_m. The average velocity during motion from one extreme point to the other extreme point will be

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a

$\frac{π}{4} v_{m}$

b

$\frac{4}{π} v_{m}$

c

$\frac{π}{2} v_{m}$

d

$\frac{2}{π} v_{m}$

answer is B.

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Detailed Solution

$v_{m} = Aω = \frac{2 πA}{T}$

$< v > = \frac{2 A}{T / 2} = \frac{4 A}{T}$

$(a v e r a g e v e l o c i t y = \frac{t o t a l d i s p l a c e m e n t}{t o t a l t i m e})$

But $\frac{A}{T} = \frac{v_{m}}{2 π}$ from Eq. (1)

$∴ < v > = 4 (\frac{v_{m}}{2 π}) = \frac{2}{π} v_{m}$

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Maximum velocity in SHM is vm. The average velocity during motion from one extreme point to the other extreme point will be