Find out the ratio of their amplituds. What are the time periods of the two motions ?

Question

Two simple harmonic motions are represented by the following equations

y₁ = 10 sin(π/4) (12t + 1)

y₂ = 5 (sin 3πt + √3 cos 3πt)

Here t is in seconds.

Find out the ratio of their amplituds. What are the time periods of the two motions ?

Answer 1

Given equations are

y₁ = 10 sin(π/4) (12t + 1) ...(1)

y₂ = 5(sin 3πt + √3 cos 3πt) ...(2)

We recast these in the form of standard equation of SHM which is

y =A sin (ωt + ϕ) ...(3)

Equation (1) maybe written as

y₁ = 10 sin[(12πt/4) + (π/4)]

y ₁ = 10 sin[(3πt + π/4)] ...(4)

Comparing eqn. (4) with eqn. (3) we have

Amplitude of first SHM = A₁ = 10 cm s ^–1 and ω₁ = 3π

∴ Time period of first motion

T₁ = 2π/ω1 = 2π/3π = (2/3) s

Eq. (2) may also be written as

y₂ = 5sin 3πt + 5 √3 cos 3πt

Let us put

i.e. amplitude of second SHM = 10 cm

and time period of second SHM = T₂ 

2π/ω² = 2π/3π = 2/3 s

Thus, the ratio of amplitudes

A₁ :A₂ = 1 : 1 and periodic times are

T₁ = T₂ = (2/3) s