Question #75d32

1 Answer
Jan 24, 2018

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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 (l-h) (from the diagram)

Now, h/l = cos theta or, h = l cos theta ,so (l-h) = l(1-cos theta)

Now, in its pathway,energy will be conserved.

So,at its mean position,total energy is kinetic energy i.e 1/2mv^2

And, at the highest point,its total energy is purely potential energy i.e mg(l-h) i.e mgl(1-cos theta)

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 omegat ......1 (here,a is the amplitude of its motion)

For,a simple pendulum, omega = sqrt(g/l),where g is acceleration due to gravity and l is the length of the pendulum)

So, its velocity equation will be, v=aomega cos omegat (by differentiating 1, as v=(dx)/dt

From 1 we can say cos omegat = sqrt(1-(x/a)^2) (as sin^2omegat+cos^2omegat=1)

So,putting the value of cosomegat in the velocity equation,we get,

v=omegasqrt(a^2-x^2) ....2

Now, for the given equation, x=asinomegat, mean position is at x=0

So,putting x=0 in equation 2 we get, v=omegaa

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 omegaa

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