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Show that the particle speed can never be equal to the wave speed in a sine wave if the amplitude is less than wavelength divided by 2π. |
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Answer» The wave equation can also be written as y = A.sin2π(t/T-x/λ) where y is the transverse displacement of the particle at a distance x at time t and A is the amplitude of the particle. The speed of the particle can be found out by partial differentiation of y with respect to t. ∂y/∂t = A.cos 2π(t/T-x/λ)*2π/T →V = {2πA.cos 2π(t/T-x/λ)}/T For V to be maximum cos2π(t/T-x/λ) = 1 So Vₘₐₓ = 2πA/T if A =λ/2π Vₘₐₓ = λ/T = v, where v is the wave speed. But if A < λ/2π, then Vₘₐₓ < λ/T →Vₘₐₓ < v. So, if the amplitude is less than the wavelength divided by 2π, the particle speed can never be more than the wave speed. |
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