# Question 75d32

Jan 24, 2018

See the diagram made,where a pendulum of string length $l$ and a bob of mass $m$ reaches up to the shown position from its mean point,and at the mean position, it had a velocity of $v$

So,vertically it has shifted a distance of $\left(l - h\right)$ (from the diagram)

Now, $\frac{h}{l} = \cos \theta$ or, $h = l \cos \theta$ ,so $\left(l - h\right) = l \left(1 - \cos \theta\right)$

Now, in its pathway,energy will be conserved.

So,at its mean position,total energy is kinetic energy i.e $\frac{1}{2} m {v}^{2}$

And, at the highest point,its total energy is purely potential energy i.e $m g \left(l - h\right)$ i.e $m g l \left(1 - \cos \theta\right)$

So,equating both we get,

v=sqrt(2gl(1-cos theta)#

ALTERNATIVELY,

Suppose, a particle in S.H.M follows the equation,

$x = a \sin \omega t$ ......1 (here,$a$ is the amplitude of its motion)

For,a simple pendulum, $\omega = \sqrt{\frac{g}{l}}$,where $g$ is acceleration due to gravity and $l$ is the length of the pendulum)

So, its velocity equation will be, $v = a \omega \cos \omega t$ (by differentiating 1, as $v = \frac{\mathrm{dx}}{\mathrm{dt}}$

From 1 we can say $\cos \omega t = \sqrt{1 - {\left(\frac{x}{a}\right)}^{2}}$ (as ${\sin}^{2} \omega t + {\cos}^{2} \omega t = 1$)

So,putting the value of $\cos \omega t$ in the velocity equation,we get,

$v = \omega \sqrt{{a}^{2} - {x}^{2}}$ ....2

Now, for the given equation, $x = a \sin \omega t$, mean position is at $x = 0$

So,putting $x = 0$ in equation 2 we get, $v = \omega a$

This the equation of the velocity of a particle under SHM at its mean position.

Now, see maximum value of $v$ in the equation 2 will come,when $x$ will be zero,so maximum value of $v$ becomes $\omega a$

That means,velocity is the maximum in the mean position as well.