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.
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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
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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
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