A 2.31-kg rope is stretched between supports that are 10.4 m apart,and has a tension in it of 49.2 N.If one end of the rope is slightly tweaked,how long will it take for the resulting disturbance to reach the other end?
A)0.699 s
B)0.720 s
C)0.664 s
D)0.615 s

Question

Quizplus · Accepted Answer

The speed of a wave on a string is given by v = sqrt(T/μ), where T is the tension and μ is the linear mass density (mass/length). Here, μ = 2.31 kg / 10.4 m = 0.222 kg/m. The speed v = sqrt(49.2 N / 0.222 kg/m) = 14.88 m/s. The time to travel 10.4 m is 10.4 m / 14.88 m/s = 0.699 s.

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