Question

From differential equation of linear S.H.M., obtain an expression for acceleration, velocity and displacement of a particle performing S.H.M.

Answer 1

The differential equation of linear SHM is 

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

where m = mass of the particle performing SHM,

`(d^2x)/dt^2=`acceleration of the particle when its displacement from the mean position is x and k = force constant.

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

Let `k/m = ω ^2`

∴ `(d^2x)/dt^2+ω ^2x=0`

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

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

`a= (d^2x)/dt^2 = (dnu)/dt=(dnu)/dx. (dx)/(dt)=(dnu)/dx.nu=nu(dnu)/dx`

Hence, Eq.(1) can be written as

`nu (dnu)/dx=-ω^2x`

`∴ nudnu=- ω^2 xdx`

Integrating this expression, we get

`nu^2/2 = (-ω ^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, ν = 0 when `x = +- A`, where A is the amplitude.

∴ `0= (-ω^2A^2)/2+C`    ` ∴ C = (ω^2 A^2)/2`

∴ `nu^2 /2 = (-ω^2x^2)/2 + (ω^2A^2)/2`

∴ `nu^2 =ω^2 (A^2 -x^2) `

∴ `nu= +- ω sqrt (A^2-X^2)`

This euation gives the velocity of the particle in terms of the displacement, x. The velocity towards right is taken to be positive and toward left as negative. Since, `nu = dx//dt` we can write Eq. (2) as follows :

`dx/dt =ω sqrt(A^2-X^2)`

∴ `dx/sqrt (A^2-X^2)=ω\ dt `

Integrating this expression, we get, 

`sin^-1 (x/A)=ωt+ alpha`...................(3)

where the constant of integration, `alpha `, is found from the initial conditions, i.e., the displacement and the velocity of the particle at time t = 0
From equation (3), we have

`x/A= sin (ωt+alpha)`

∴ Displacement as a function of time is,

`x = A sin (ωt + alpha)`