Question

From differential equation of linear simple harmonic motion obtain an expression for velocity of a particle performing linear simple harmonic motion

Answer 1

Answer:

Hence obtain the expression for acceleration, velocity and displacement of a particle performing linear S.H.M. The differential equation of linear SHM is d2→xdt2+km→x=0 d 2 x → d t 2 + k m x → = 0 where m = mass of the particle performing SHM. ∴ Acceleration, a = d2xdt2=−ω2x d 2 x d t 2 = - ω 2 x .... (1).

Answer 2

Answer:

Velocity is the rate of change of displacement. The expression for velocity can be obtained from the expression of acceleration.

Acceleration, 

dt2d2x=dtdv=dxdv×dtdx

But, acceleration =v(dtdv)−−−−−(1)

But we know, dt2d2x=−ω2x----- (2)

From (1) and (2), vdxdv=−ω2x

∴vdv=−ω2x

Integrating both side, the above equation, we get

∫vdv=∫−ω2xdx=−ω2∫xdx

Hence, 2v2=2−ω2k2+c

where C is the constant of integration. Now, to find the vaue of C, lets consider boundary value condition. When a particle performing SHM is at the extreme position, displacement of the particle is maximum and velocity is zero. 

a is the amplitude of SHM.

Therefore, at x=±a,v=0

&0=2−ω