|Class 11 Physics Oscillations||Phase|
It is that quantity that determines the state of motion of the particle.
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)
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...