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:
\(\frac{B\,sin\,a}{B\,sin\,a}=\frac{1}{1}\)
\(tan\,a=1=tan\,\frac{\pi}{4}\)
\(∴a=\frac{\pi}{4},\frac{5\pi}{4},.......\)