Question

Equation of a transverse wave travelling in a rope is given by y=5sin(4.0t-0.02 x) where y and x are expressed in cm and time in seconds. Calculate (a) the amplitude, frequency,velocity and wavelength of the wave. (b) the maximum transverse speed and acceleration of a particle in the rope.

Answer 1

On comparing with standard equation i.e.

Y=Asin(wt-kx)

We have A(amplitude):5

Frequency:w/2π=.6369

Velocity: w/k =200

Wavelength:2π/k=314m

Maximum transverse speed:wA=20 m/sec

Maximum acceleration:Aw2=40m/sec^2

Answer 2

The maximum transverse speed is 20 cm/s and maximum transverse acceleration is -80 cm/s^2

Explanation:

(a) Comparing this with the standard equation of wave motion,

y = Asin(ωt − kx) = A  sin(2πft − 2π / λ.x)

Where A, f and lambda are amplitude, frequency and wavelength respectively.

Thus, amplitude A=5 cm

⇒ 2πf = 4

⇒ Frequency "f" = (4) / (2π) = 0.637 Hz

v = f λ

v = ω / K = 4 / 0.02 = 200 cm/s = 2 m/s

λ = 2/2 π => πm

(b) Maximum transverse speed =  ωA = 4 x 5 = 20 cm / s

Maximum transverse acceleration =  - ωA^2 =  - 16 x 5 = - 80 cm/s^2

Hence the maximum transverse speed is 20 cm/s and maximum transverse acceleration is -80 cm/s^2