25. (a) The displacement of the string is assumed to have the form y ( x, t )= y m sin( kx − ωt ). The velocity of a point on the string is u ( x, t )= ∂y/∂t = − ωy m cos( kx − ωt ) and its maximum value is u m = ωy m . For this wave the frequency is f = 120 Hz and the angular frequency is ω =2 πf =2 π (120 Hz) = 754 rad / s. Since the bar moves through a distance of 1 . 00 cm, the amplitude is half of that, or y m =5 . 00 × 10 − 3 m. The maximum speed is u m = (754 rad / s)(5 . 00 × 10 − 3 m) = 3 . 77 m / s. (b) Consider the string at coordinate x and at time t and suppose it makes the angle θ with the x axis. The tension is along the string and makes the same angle with the x axis. Its transverse component is τ trans = τ sin θ .Now θ is given by tan θ = ∂y/∂x = ky m cos( kx − ωt ) and its maximum value is given by tan θ m = ky m . We must calculate the angular wave number k .I ti sg i v e nb y
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