Question

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.

Answer 1

Initially, at t = 0:
Displacement, x = 1 cm
initial velocity, v = ω cm/sec.
Angular frequency, ω = π rad/s –1
It is given that x(t)= A cos(ω t+ϕ)
1=A cos(ω x 0+ϕ)= A cos ϕ
A cos ϕ=1 (i)
velocity, v=$\frac{dx}{dt}$
ω=-A ωsin(ωt+ϕ)
1=A sin(ωt+0+ϕ)=- A sin ϕ)
A sin ϕ=-1 (ii)
Squaring and adding equations (i) and (ii), we get:
A2 (Sin2 ϕ+cos2 ϕ)=1+1
A2 =2
∴ ϕ=$\frac{3\pi}{4},\frac{7\pi}{4},.....$
SHM is given as:
x= B sin (ωt+α)
Putting the given values in this equation, we get:
1=B sin[ωt x 0+α]
B sin α=1 (iii)
Velocity, v= ωB cos (ωt+α)
Substituting the given values, we get:
$\pi=\pi\,B\,sin\,α$
B sin α=1
Squaring and adding equations (iii) and (iv), we get
$B^2[sin^2a+cos^2a]=1+1$
$B^2=2$
$∴ B=\sqrt2\,cm$
Dividing equation (iii) by equation (iv), we get:
$\frac{B\,sin\,a}{B\,sin\,a}=\frac{1}{1}$
$tan\,a=1=tan\,\frac{\pi}{4}$
$∴a=\frac{\pi}{4},\frac{5\pi}{4},.......$