The motion of a particle executing simple harmonic motion is described by the displacement function,

x(t) = A cos (ωt + φ).

If the initial (t = 0) position of the particle is 1 cm and its initial velocity is ω cm/s, what are its amplitude and initial phase angle? The angular frequency of the particle is π s^–1. If instead of the cosine function, we choose the sine function to describe the SHM x = B sin (ωt + α), what are the amplitude and initial phase of the particle with the above initial conditions.

We have the given displacement function as

x(t) = A cos (ωt + φ ) …….(1)

The initial conditions are,

t = 0, x(0) = 1 cm i.e the position of the particle and angular frequency, ω = π s^–1

Substituting the initial conditions in the displacement function in equation(1)

Hence, 1 = A cos(π .0+ φ )

A cos φ = 1 ………(2)

Differentiating (1) w.r.t ‘t’

The derivative of displacement is velocity.

Hence,

v = -Aωsin(ωt + φ ) …………(3)

At t = 0, v = ω (Initially)

Hence, from (3) we get,

ω = -Aωsin(π .0 + φ )

Also, Asinφ = -1 …………….(4)

Now we square and add (1) and (4)

We get A = √2 cm

Dividing (2) and (4),

Asin φ/Acos φ = -1/1

Hence, φ = 3π/4

We use the sine function

x = B sin (ωt + α)

we get Bsin α = 1 …..(5)

at t = 0, using x = 1 and v = ω

and Bsin α = 1 ……….(6)

Dividing (5) and (6)

tan α = 1 and α = π/4 and 5π/4

Squaring and adding (5) and (6) we get

B = √2 cm.