## College Physics (4th Edition)

The pulse will move faster on the string with a mass of 58.0 grams. The slower pulse will require an additional $~6.7~ms~$ to reach the end of its string.
Let $v_1$ be the speed of a pulse on the string of mass $78.0~g$. We can find $v_1$: $v_1 = \sqrt{\frac{F}{m/L}}$ $v_1 = \sqrt{\frac{180.0~N}{78.0\times 10^{-3}~kg/15.0~m}}$ $v_1 = 186~m/s$ Let $v_2$ be the speed of a pulse on the string of mass $58.0~g$. We can find $v_2$: $v_2 = \sqrt{\frac{F}{m/L}}$ $v_2 = \sqrt{\frac{160.0~N}{58.0\times 10^{-3}~kg/15.0~m}}$ $v_2 = 203~m/s$ The pulse will move faster on the string with a mass of 58.0 grams. We can find the time $t_1$ for the slower pulse to reach the end of its string: $t_1 = \frac{L}{v_1} = \frac{15.0~m}{186~m/s} = 0.0806~s$ We can find the time $t_2$ for the faster pulse to reach the end of its string: $t_2 = \frac{L}{v_2} = \frac{15.0~m}{203~m/s} = 0.0739~s$ We can find the time difference: $\Delta t = t_1-t_2 = (0.0806~s)-(0.0739~s) = 0.0067~s = 6.7~ms$ The slower pulse will require an additional $~6.7~ms~$ to reach the end of its string.