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 (lh) (from the diagram)

Now, hl=cosθ or, h=lcosθ ,so (lh)=l(1cosθ)

Now, in its pathway,energy will be conserved.

So,at its mean position,total energy is kinetic energy i.e 12mv2

And, at the highest point,its total energy is purely potential energy i.e mg(lh) i.e mgl(1cosθ)

So,equating both we get,

v=2gl(1cosθ)

ALTERNATIVELY,

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

x=asinωt ......1 (here,a is the amplitude of its motion)

For,a simple pendulum, ω=gl,where g is acceleration due to gravity and l is the length of the pendulum)

So, its velocity equation will be, v=aωcosωt (by differentiating 1, as v=dxdt

From 1 we can say cosωt=1(xa)2 (as sin2ωt+cos2ωt=1)

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

v=ωa2x2 ....2

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

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

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 ωa

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