NCERT Solutions
Class 11 Physics
Oscillations
Q. 13.7
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...