Question

Deduce an expression for the velocity of a particle executing shm when particle velocity at maximum and minimum

Answer 1

hope it helps!
please mark as brainlist.

Answer 2

Maximum velocity is $V_{max}=\omega \cdot r$ and Minimum velocity $V_{min}=0$.

Explanation:

• Suppose that a particle is in SHM of amplitude

r = movement of a particle at time t is

$y = r\hspace{0.05 cm}sin\hspace{0.05 cm}\omega t$

The velocity of the particle at a time t is given by

$v=\frac{dy}{dt}$

$v=\frac{d}{dt}(r\hspace{0.05 cm}sin\hspace{0.05 cm}\omega t)$

$v=(r\hspace{0.05 cm}\omega\hspace{0.05 cm}cos\hspace{0.05 cm}\omega t)$

$v=r\hspace{0.05 cm}\omega\hspace{0.05 cm}\sqrt{1-sin^{2}\hspace{0.05 cm}\omega\hspace{0.05 cm}t}$

$v=r\hspace{0.05 cm}\omega\hspace{0.05 cm}\sqrt{1-\frac{y^{2}}{r^{2}}}$

$v=r\hspace{0.05 cm}\omega\hspace{0.05 cm}\sqrt{\frac{r^{2}-y^{2}}{r^{2}}}$

$v=\frac{r\hspace{0.05 cm}\omega}{r}\hspace{0.05 cm}\sqrt{r^{2}-y^{2}}}$

$v=\hspace{0.05 cm}\omega\hspace{0.05 cm}\sqrt{r^{2}-y^{2}}}$

Thus, particle is at one mean position then,

$y=0$

So,

$v=\hspace{0.05 cm}\omega\hspace{0.05 cm}\sqrt{r^{2}-y^{2}}}$

put y=0

$v=\hspace{0.05 cm}\omega\hspace{0.05 cm}\sqrt{r^{2}-0}}$

$v=\hspace{0.05 cm}\omega\hspace{0.05 cm}\sqrt{r^{2}}}$

$v=\hspace{0.05 cm}\omega\hspace{0.05 cm}r$ (Max velocity)

• It is also called velocity amplitude for SHM.
• If the particle is at an extreme position then,

$y=r$

So,

$v=\hspace{0.05 cm}\omega\hspace{0.05 cm}\sqrt{r^{2}-y^{2}}}$

put y=r

$v=\hspace{0.05 cm}\omega\hspace{0.05 cm}\sqrt{r^{2}-r^{2}}}$

$v=\hspace{0.05 cm}\omega\hspace{0.05 cm}\sqrt{0}}$

$v=0$ (Min velocity)

Thus, Maximum velocity is $V_{max}=\omega \cdot r$ Minimum velocity $V_{min}=0$.