SHOW that when a particle is moving in SHM its velocity at a distance of root 3 by 2 A from its mean position is half of its velocity at mean position (where, A = amplitude of SHM)
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Hello there,
● Proof -
Let's consider a particle moving in SHM with angular vekocity ω and amplitude A.
Velocity of the particle at position x from mid-point is given by -
v = ω √(A^2-x^2)
When particle is at x = √3/2 A
v1 = ω √(A^2 - 3/4 A^2)
v1 = Aω/2 ...(1)
When particle is at mean position -
v2 = ω √(A^2 - 0)
v2 = Aω ...(2)
From (1) & (2),
v1 = v2 / 2
That is, for a particle in SHM, velocity at a distance of √3/2 A from its mean position is half of its velocity at mean position.
hope this helps...
● Proof -
Let's consider a particle moving in SHM with angular vekocity ω and amplitude A.
Velocity of the particle at position x from mid-point is given by -
v = ω √(A^2-x^2)
When particle is at x = √3/2 A
v1 = ω √(A^2 - 3/4 A^2)
v1 = Aω/2 ...(1)
When particle is at mean position -
v2 = ω √(A^2 - 0)
v2 = Aω ...(2)
From (1) & (2),
v1 = v2 / 2
That is, for a particle in SHM, velocity at a distance of √3/2 A from its mean position is half of its velocity at mean position.
hope this helps...
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