Question

derive equation for velocity acceleration and time period of a particle executing simple harmonic motion​

Answer 1

Answer:

Explanation:

can obtain the expression for velocity using the expression for acceleration.Let’s see how. Acceleration d2x/dt2 = dv/dt = dv/dx × dx/dt. But dx/dt = velocity ‘v’

Therefore, acceleration = v(dv/dx)                                                                (II)

When we substitute equation II in equation I, we get, v(dv/dx) =  – ω2 x.

∴ vdv = – ω2 xdx

After integrating both sides, we get,

∫vdv = ∫-ω2 xdx = -ω2∫ xdx

Hence, v2 /2 = -ω2 x2 /2 + 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

And 0 = – ω2 a2 /2 + C

Hence, C = ω2 a2 /2

Let’s substitute this value of C in equation v2 /2 = -ω2x2 /2 + C

∴ v2 /2 = -ω2 x2 /2 + ω2 a2 /2

∴  v2 = ω2  ( a2  – x2  )

Taking square root on both sides, we get,

v = ± ω  √(a2  – x2 )                                                                                 (III)

Equation III is the expression of the velocity of S.H.M. The double sign indicates that when a particle passes through a given point in the positive direction of x, v is positive, and when it passes through the same point in opposite direction of x, v is negative.

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