A 1.70-m string of weight 1.00 N is tied to the celling at its upper end, and th

Question

A 1.70-m string of weight 1.00 N is tied to the celling at its upper end, and the lower end supports a weight W. when you pluck the string slightly, the waves traveling up the string obey the equation y(x, t) = (8.75 mm) cos(180m^-1x = 2760s^-1 t). (a) How much time does it take a pulse to travel the full length of the string? 11 s (b) What is the weight W? 13.77 N (c) How many wavelengths are on the string at any instant of time? (d) What is the equation for waves traveling down the string? y(x, t) = (8.75 mm) cos(180 rad/m]x + [2760] rad/s]t)

Explanation / Answer

L = 1.7 m ; W = 1 N ;

y(x,t) = (8.75 mm) cos(180 m^-1 x -2760 s^-1 t)

a)v = w/k

v = 2760/180 = 15.33

t = L/v = 1.7/15.33 = 0.11 s

Hence, t = 0.11 s

b)v = sqrt (F/u)

v^2 = F/u => F = u v^2

u = m/L

F = (m/L) v^2

F = (w/g)v^2/L = w v^2/g L

F = 1 x 15.33^2/1.7 x 9.8 = 14.1 N

Hence, F = 14.1 N

c)L/lambda = (1/2 pi)(2 pi/lambda)L

lambda = k L/2 pi = 180 x 1.7/2 x 3.14 = 48.73 m

Hence, lambda = 48.73 m

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