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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.

User Tuxnani
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Answer

given,


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

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


f = (1)/(\tau)


f = (1)/(3.9 * 10^(-2))

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.

User Alfred Xiao
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