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

##### 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)

1=Acos(ω×0+ϕ)=Acosϕ

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

1=−Asin(ω×0+ϕ)=−Asinϕ

Asinϕ=−1 ———-(4)

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

A2(sin2ϕ+cos2ϕ)=1+1

A = $$\sqrt{2}$$ cm

Dividing equation (4) by equation (2),

$$\dfrac{A sin ϕ}{A cos ϕ}=\dfrac{-1}{1}$$

tanϕ=−1

⇒ ϕ=$$\dfrac{3π}{4},\dfrac{7π}{4}$$

If sine function is used

x=Bsin(ωt+α)

At t = 0, x = 1 and v = ω we get

1=Bsin(ω×0+α]=1+1

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:

B2[sin2α+cos2α]=1+1

B=2

B= $$\sqrt{2}$$ cm

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

$$\dfrac{Bsinα}{Bcosα}=\dfrac{-1}{1}$$

tanα=1

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

Answered by Pragya Singh | 1 year ago

### Related Questions

#### A mass attached to a spring is free to oscillate, with angular velocity ω, in a horizontal plane without friction

A mass attached to a spring is free to oscillate, with angular velocity ω, in a horizontal plane without friction or damping. It is pulled to a distance x0 and pushed towards the centre with a velocity v0 at time t = 0. Determine the amplitude of the resulting oscillations in terms of the parameters ω, x0 and v0. [Hint: Start with the equation x = a cos (ωt+θ) and note that the initial velocity is negative.]

#### A body describes simple harmonic motion with an amplitude of 5 cm and a period of 0.2 s.

A body describes simple harmonic motion with an amplitude of 5 cm and a period of 0.2 s. Find the acceleration and velocity of the body when the displacement is

(a) 5 cm

(b) 3 cm

(c) 0 cm.

#### A circular disc of mass 10 kg is suspended by a wire attached to its centre.

A circular disc of mass 10 kg is suspended by a wire attached to its centre. The wire is twisted by rotating the disc and released. The period of torsional oscillations is found to be 1.5 s. The radius of the disc is 15 cm. Determine the torsional spring constant of the wire. (Torsional spring constant α is defined by the relation J = –α θ, where J is the restoring couple and θ the angle of twist).

#### Show that for a particle in linear SHM the average kinetic energy over a period of oscillation equals.

Show that for a particle in linear SHM the average kinetic energy over a period of oscillation equals the average potential energy over the same period.