Question

The displacement of an oscillating particle is given by x = a sin ω t + b cos ω t where a, b and ω are constants. Prove that the particle performs a linear S.H.M. with amplitude A = `sqrt("a"^2 + "b"^2)`.

Answer 1

Using trigonometry,

From the above diagram,

`a/sqrt(a^2+b^2)=costheta, "and" b/sqrt(a^2+b^2)=sintheta`

Displacement of a particle performing oscillation,

`x = asinomegat+bcosomegat`

`= sqrt(a^2+b^2)[a/sqrt(a^2+b^2)sinomegat+b/sqrt(a^2+b^2)cosomegat]`

`=sqrt(a^2+b^2)[costhetasinomegat+sinthetacosomegat]`

`=sqrt(a^2+b^2) sin (omegat+theta)`

On comparing it with general equation of SHM

`x = Asin(omegat+Φ)` we get

Amplitude as, `A =sqrt(a^2+b^2) `

Answer 2

The displacement of the particle is x = a sin(ωt) + b cos(ωt).

The velocity of the particle is:

v = `"dx"/"dt"`

⇒ `v = "d"/"dt" (a sin (omega "t") + b cos (omega "t"))`

⇒ v = a ω cos (ωt) - b ω sin (ωt)

The acceleration of the particle is:

a = `"dv"/"dt"`

⇒ `a = "d"/"dt"` (a ω cos (ωt) - b ω sin (ωt))

⇒ a = - aω² sin (ωt) - bω² cos (ωt)

⇒ a = - ω² x

Since the acceleration of the particles is directly proportional to displacement towards the mean position. So the motion is SHM.

Let the amplitude be a = A cos (Φ), b = A sin (Φ)

Substituting these values in x = a sin (ωt) + b cos (ωt)

x = A cos (Φ) sin (ωt) + A sin (Φ) cos (ωt)

⇒ x = A sin (ωt + Φ)

Squaring the amplitudes and adding them, 

⇒ a² + b² = A² cos² (Φ) + A² sin² (Φ)

⇒ a² + b² = A² 

⇒ A = `sqrt("a"^2 + "b"^2)`