Transverse wave

Question

A transverse wave is described by the expression: y= 0.05cos(4.19x - 1260t)

a) What is the amplitude of this wave?

b) What is the wavelength of this wave?

c) What is the frequency of this wave?

d) How fast is this wave travelling?

e) What is the maximum transverse velocity of this wave?

Answer 1

Compare the given equation to the general equation i.e y = Acos(2π(x/λ - t/T)), where A = amplitude.

a) amplitude, A = 0.05m

b) wavelength, λ = 2π/4.19 = 1.5m

c) frequency, f = 1/T =1260/2π = 1979.2 Hz

d) velocity, v = λf = 2968.8 m/s

e) the maximum velocity is v_max = Aω = (0.05m(1260rad/s) = 63m/s [this is obtained by differentiating y with respect to time, t]