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A body is in simple harmonic motion with time period half second (T=0.5 s) and amplitude one cm (A=1 cm). Find the average velocity in the interval in which it moves from equilibrium position to half of its amplitude.

Option 1)
$4 \: cm/s$

Option 2)
$6\: cm/s$

Option 3)
$12 \: cm/s$

Option 4)
$16 \: cm/s$

Answers (1)

best_answer

As we learned

Equation of S.H.M. -

$a=-\frac{d^{2}x}{dt^{2}}= -w^{2}x$

$w= \sqrt{\frac{k}{m}}$

- wherein

$x= A\sin \left ( wt+\delta \right )$

$A=A_{0}\sin wt$

at $A=\frac{A_{0}}{2}\Rightarrow sinwt=\frac{1}{2}$

$\therefore wt=\frac{\pi }{6}\Rightarrow t=\frac{\pi }{6w}$

$t=\frac{\pi }{6\times \frac{2\pi }{\tau }}=\frac{\tau }{12}=\frac{1}{24}sec$

$\upsilon =Aw\cdot \cos wt$

$\upsilon _{av}=\frac{Aw_{0}\int^{\tau /12}\cos wt dt}{^{\tau /12}\int_{0} }=Aw\cdot \frac{\sin wt\int_{0}^{\tau /12}}{w\cdot \frac{\tau }{12}}$

$=Aw\cdot \frac{\sin \left ( \frac{2\pi }{\not{\tau }} \cdot \frac{\pi _(\not{\tau })}{12}\right )}{\left ( \frac{w\tau }{12} \right )}=\frac{Aw\cdot (1/2)}{\frac{2\pi }{12}}$

$=A\cdot \left ( \frac{\not{2\pi }}{\tau } \right )\cdot \frac{1}{2}\cdot \frac{12}{\not{2\pi }}=\frac{A\times 12}{0.5\times 2}$

=12A

$V_{av}=12cm/s$

Option 1)

$4 \: cm/s$

Option 2)

$6\: cm/s$

Option 3)

$12 \: cm/s$

Option 4)

$16 \: cm/s$

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