Question

Obtain the differential equation of linear simple harmonic motion.

Derive the differential equation of linear S.H.M. 

Answer 1

Let a particle of mass 'm’ undergo S.H.M. about its mean position O. At any instant ‘t’, displacement of the particle will be ‘x’ as shown in the following figure.

By definition, F = −kx            ...(1) 

where k is force constant

The acceleration of the particle is given by,

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

According to Newton’s second law of motion,

F = ma

`therefore F=m((d^2x)/dt^2)`            ...(2)

From equations (1) and (2),

`m((d^2x)/dt^2)=-kx`

`therefore (d^2x)/dt^2=-k/mx`

`therefore (d^2x)/dt^2+k/mx=0`                ...(3)

where, `k/m=omega^2="constant"`

`therefore (d^2x)/dt^2+omega^2x=0`                   ...(4)

Equations (3) and (4) represent differential equations of linear S.H.M.

Answer 2

(i) In a linear S.H.M., the force is directed towards the mean position, and its magnitude is directly proportional to the displacement of the body from the mean position.

∴ f ∝ −x

∴ f = −kx     ...(1)

where k is the force constant and x is the displacement from the mean position.

(ii) According to Newton’s second law of motion, 

f = ma    ….(2)

From equations (1) and (2), 

ma = –kx    ….(3)

(iii) The velocity of the particle is given by `"v" = "dx"/"dt"`

∴ Acceleration, a = `"dv"/"dt" = ("d"^2"x")/"dt"^2`     ...(4)

Substituting equation (4) in equation (3),

`"m" ("d"^2"x")/"dt"^2 = -"kx"`

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

(iv) Substituting `"k"/"m" = omega^2`, where ω is the angular frequency,

∴ `("d"^2"x")/"dt"^2 + omega^2"x" = 0`

This is the differential equation of linear S.H.M.