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.

Asked by Pragya Singh | 1 year ago |  130

1 Answer

Solution :-

Displacement function x(t) = A cos (ωt + φ )

At position,  t = 0;

Displacement, x(0)=1 cm

Initial velocity, v=ω cm/s

Angular frequency, ω=π s−1

The given function is  x(t)=Acos(ωt+ϕ) ——–(1)


Acosϕ=1  —————(2)

Differentiating equation (1) w.r.t “t”

Velocity, v=\( \dfrac{dx}{dt}\)

v=−Aωsin(ωt+ϕ) ——(3)

t = 0 and v = ω


Asinϕ=−1 ———-(4)

Squaring and adding equations (2) and (4), we get:


A = \( \sqrt{2}\) cm

Dividing equation (4) by equation (2),

\( \dfrac{A sin ϕ}{A cos ϕ}=\dfrac{-1}{1}\)


⇒ ϕ=\( \dfrac{3π}{4},\dfrac{7π}{4}\)

If sine function is used


At t = 0, x = 1 and v = ω we get


Bsinα=1 —- (5)

Velocity, v= \( \dfrac{dx}{dt}\) = Bωcos(ωt+α)

taking v = ω

1 = Bcos(ω(0)+α) = Bcos α——(6)

Squaring and adding equations(5) and (6), we get:



B= \( \sqrt{2}\) cm

Dividing equation (5) by equation (6), we get:

\( \dfrac{Bsinα}{Bcosα}=\dfrac{-1}{1}\)


Therefore, α=\( \dfrac{π}{4},\dfrac{5π}{4}\),……

Answered by Pragya Singh | 1 year ago

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