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

Given,
X = Acos(wt + ø)
at t = 0 position of particle is 1 cm and velocity = w cm/s

so , 1 = Acos( w×0 + ø)
1 = Acosø ------(1)

again,
x = Acos(wt + ø)
differentiate wrt t
dx/dt = -Awsin(wt + ø)
at t = 0 , V = w cm/s
So, w = -Awsin(w×0 +ø)
1 = -Asinø ------(2)
From eqns (1) and(2)
2 = A²( cos²ø + sin²ø)
A = ±√2
Hence, amplitude {A} = √2 cm
Again,
Divide eqns (2) ÷ (1)
tanø = -1
ø = -π/4 or 7π/4


Now ,if we choose
X = Bsin(wt + a )
Then, at t = 0 ,
X = 1 cm
V = w cm/s
Put t = 1 and x = 1 in above equation .
1 = Bsina ------(1)
and
dx/dt = Bwcos(wt +a)
Put t = 0 and V = w
1 = Bcosa -----(2)
From eqns (1) and (2)
B = ±√2
Hence amplitude = √2 cm
Tana = 1
a = π/4

Answer 2

Answer:

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