NCERT Solutions
Class 11 Physics
Oscillations

Q. 13.12
Plot the corresponding reference circle for each of the following simple harmonic motions. Indicate the initial (t =0) position of the particle, the radius of the circle, and the angular speed of the rotating particle. For simplicity, the sense of rotation may be fixed to be anticlockwise in every case: (x is in cm and t is in s).
(a) x = –2 sin (3t + π/3)
(b) x = cos (π/6 – t)
(c) x = 3 sin (2πt + π/4)
(d) x = 2 cos πt
- x = -2 sin(3t + (π/3))
= + 2 cos (3t + (π/3) + (π/3))
=2 cos (3t + ((5π/6))
Comparing the above equation with the standard SHM equation
x = A cos ((2 πt)/T + ϕ), then
Amplitude A = 2cm
Phase angle ϕ = (5π/6) = 1500
Angular velocity ω, = (2π/T) = 3 rad/sec
The motion of the particle can be plotted as shown in the following figure.
x= cos ((π/6) – t) = cos (t – (π/6))
Comparing the above equation with the standard SHM equation
x = A cos ((2 πt)/T + ϕ), then
Amplitude A = 1cm
Phase angle ϕ= - (π/6) = -300
Angular velocity ω = (2π/T) = 1 rad/sec
The motion of the particle can be plotted as shown in the following figure.
x= 3 sin ((2π/T) + (π/4))
= -3 cos [(2πt + (π/4)) + (π/2)]
=-3cos (2πt + (3π/4))
Comparing the above equation with the standard SHM equation
x = A cos ((2 πt)/T + ϕ), then
Amplitude, A = 3 cm
Phase angle, ϕ = (3 π /4) = 135°
Angular velocity, ω = (2π/T) = 2π rad/sec
The motion of the particle can be plotted as shown in the following figure.
Comparing the above equation with the standard SHM equation
x = A cos ((2 πt)/T + ϕ), then
Amplitude, A = 2 cm
Phase angle, Φ = 0
Angular velocity, ω = π rad/s
The motion of the particle can be plotted as shown in the following figure