Chapter 16, Problem 052 A rope, under a tension of 259 N and fixed at both ends,

Question

Chapter 16, Problem 052 A rope, under a tension of 259 N and fixed at both ends, oscillates in a second-harmonic standing wave pattern. The displacement of the rope is given by y = (0.228m)sin(mx / 5.00)sin( 12.0m) where X 0 at one end of the rope, x is in meters, and t is in seconds. What are (a) the length of the rope, (b) the speed of the waves on the rope, and (c) the mass of the rope? (d) If the rope oscillates in a third-harmonic standing wave pattern, what will be the period of oscillation? (a) Number (b) Number (c) Number (d) Number Units Units Units Units

Explanation / Answer

Given,

T = 259 N ;

y = (0.228 m) sin (pi x/5) sin(12 pi t)

comparing with the general eqn

y(x,t) = 2 A sin(kx) cos(wt)

a)On comparing we get

k = pi/5

2 pi/lambda = pi/5 => lambda = 10 m

Hence, L = 10 m

b)w = 2 pi f

2 pi f = 12 pi => f = 6 Hz

v = f lambda = 6 x 10 = 60 m/s

Hence, w = 60 m/s

c)we know that

v = sqrt (T/(m/L))

60 = sqrt (259/m/10) => 3600 = 259 x 10/m

m = 0.72 kg

d)f = 3v/2 L

f = 3 x 60/2 x 10 = 9 Hz

T = 1/f = 1/9 = 0.11 s

Hence, T = 0.11 s

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