Class 11 Physics Oscillations Phase


It is that quantity that determines the state of motion of the particle.

  • Its value is (ωt + Φ)
  • It is dependent on time.

Value of phase at time t=0, is termed as Phase Constant. When the motion of the particle starts it goes to one of the extreme position at that time phase is considered as 0.

Let x (t) = A cos (ωt) where we are taking (Φ = 0)

  1. Mean Position (t= 0)
  2. x (0) = A cos (0) = A (cos0=1)
  3. t=T/4, t= T/2, t=3T/4, t=T and t=5T/4


The above figures depict the location of the particle in SHM at different values of t=0,T/4,T/2,3T/4,T,5T/4.The time after which motion repeats is T.The speed is maximum for zero displacement(x=0) and zero at the extremes of motion.

In the above graph the displacement as a function of time is obtained when φ = 0.The curves (1) and (2) are of two different amplitudes A and B.


In the above graph the curves (3) and (4) are for φ = 0 and -π/4 respectively but the amplitude is same for both.

Problem: - Which of the following relationships between the acceleration (a) and the displacement x of a particle involve simple harmonic motion?

(a) a = 0.7x

(b) a = –200x2

(c) a = –10x

(d) a = 100x3


In SHM, acceleration a is related to displacement by the relation of the form

 a = -kx, which is for relation (c).

Problem: - 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.


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 (ωt + Φ)

1 = A cos (ω × 0 + Φ) = A cos Φ

A cosΦ = 1                          ... (i)

Velocity, v= dx/dt

ω = -A ω sin (ωt + Φ)

1 = -A sin (ω × 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

∴ A = √2 cm

Dividing equation (ii) by equation (i), we get:

tan Φ = -1

∴Φ = 3π/4, 7π/4...

SHM is given as:

x = B sin (ωt + α)

Putting the given values in this equation, we get:

1 = B sin [ω × 0 + α] = 1 + 1

B sin α = 1                           ... (iii)

Velocity, v = ω B cos (ωt + α)

Substituting the given values, we get:

π = π B sin α

B sin α = 1                            ... (iv)

Squaring and adding equations (iii) and (iv), we get:

B2 [sin2 α + cos2 α] = 1 + 1

B2 = 2

∴ B = √2 cm

Dividing equation (iii) by equation (iv), we get:

B sin α / B cos α = 1/1

tan α = 1 = tan π/4

∴α = π/4, 5π/4...

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