Question

Hence obtain the expression for acceleration, velocity and displacement of a particle performing linear S.H.M.

Answer 1

Hence obtain the expression for acceleration, velocity and displacement of a particle performing linear S.H.M.

The differential equation of linear SHM is `(d^2vecx)/(dt^2) + k/m vecx = 0` where m = mass of the particle performing SHM.

(d^2vecx)/(dt)^2  = acceleration of the particle when its displacement from the mean position is `vecx` and k =  force constant. For linear motion, we can write the differential equation in scalar form :

`(d^2x)/(dt^2) + k/m x = 0`

Let `k/m = omega^2` , a constant

`:. (d^2x)/(dt^2) + omega^2x = 0`

∴ Acceleration, a = `(d^2x)/(dt^2) = -omega^2x` ....(1).

The minus sign shows that the acceleration and the displacement have opposite directions. Writing
v = `(dx)/(dt)` as the velocity of the particle.

`a = (d^2x)/(dt^2) = (dv)/(dt) = (dv)/(dx)xx(dx)/(dt) = (dv)/(dx) = v = v (dv)/(dx)`

Hence, Eq. (1) can be written as

`v(dv)/(dx) = -omega^2x`

∴ `vdv = -omega^2x`

Integrating this expression, we get

`v^2/2 = (-omega^2x^2)/2 + C`

where the constant of integration C is found from a boundary condition.

At an extreme position (a turning point of the motion), the velocity of the particle is zero. Thus, v = 0 when x = ±A, where A is the amplitude.

`:. 0 = (-omega^2A^2)/2 + C` `:. C = (omega^2A^2)/2`

`:.v^2/2 = (-omega^2x^2)/2 + (omega^2A^2)/2`

`:. v^2 = omega^2(A^2 -x^2)`

`:.v = +- omegasqrt(A^2 - X^2)` ......(2)

This equation gives the velocity of the particle in terms of the displacement, x. The velocity towards right is taken to be positive and towards left as negative.

Since v = dx/dt we can write Eq. 2 as follows:

`(dx)/(dt) = omegasqrt(A^2 - x^2)`(considering only the plus sign)

`:. (dx)/(sqrt(A^2-x^2)) = omega dt`

Integrating the expression, we get,

`sin^(-1) (x/A) = omegat+x`  ....(3)

where the constant of integration, x , is found from the initial conditions, i.e., the displacement and the
velocioty of the particle at time t = 0.

From Eq (3), we have

`x/A= sin(omegat + x)`

∴ Displacement as a function of time is, x = Asin (ωt + x).