# Chapter 11 - Problems - Page 428: 13

It takes $16.0~ms$ for the wave pulse to travel to the upper end.

#### Work Step by Step

We can find the linear mass density: $\mu = \frac{m}{L}$ $\mu = \frac{W/g}{L}$ $\mu = \frac{W}{g~L}$ $\mu = \frac{0.25~N}{(9.80~m/s^2)(10.0~m)}$ $\mu = 2.55\times 10^{-3}~kg/m$ Since the weight of the string is negligible, we can let $F = 1000~N$. We can find the speed of the wave pulse: $v = \sqrt{\frac{F}{\mu}}$ $v = \sqrt{\frac{1000~N}{2.55\times 10^{-3}~kg/m}}$ $v = 626~m/s$ We can find the time it takes the wave pulse to reach the upper end: $t = \frac{L}{v} = \frac{10.0~m}{626~m/s} = 0.0160~s = 16.0~ms$ It takes $16.0~ms$ for the wave pulse to travel to the upper end.

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