A certain transverse wave is described by y(x,t)=Bcos[2π(xL−tτ)],where B = 6.40 mm , L = 26.0 cm , and τ = 3.90×10^−2 s .a. Determine the wave's amplitude.b. Determine the wave's wavelength.c. Determine the wave's frequency.d. Determine the wave's speed of propagation. e. Determine the wave's direction of propagation.

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Answer

given,

[tex]y(x,t)=B cos[2\pi (\dfrac{x}{L} - \dfrac{t}{\tau})][/tex]

B = 6.40 mm ,  L = 26 cm ,    τ = 3.90 × 10⁻² s

general wave equation

  y = A cos (k x - ωt)

where A is the amplitude of the

a) Amplitude of the given wave

      B = 6.40 mm

b) Wavelength of the given wave

       λ = L

       λ = 26 cm

c)  wave frequency

     [tex]f = \dfrac{1}{\tau}[/tex]

     [tex]f = \dfrac{1}{3.9 \times 10^{-2}}[/tex]

            f = 25.64 Hz

d) speed of wave will be equal to

     v = f λ

     v = 25.64 x 0.26

     v = 6.67 m/s

e) direction of propagation will be along +ve x direction because sign of k x and ωt is same as general equation.