Chapter 11, Problem 85GP

To determine

To find: The distance where two pulses first meet.

Answer to Problem 85GP

The distance where two pulses meet is $6.44\text{m}$ .

Explanation of Solution

Given:

Length of wire, $l=\text{10}\text{.0m}$

Tension, $t=\text{255N}$

Mass of wire, $m=\text{0}\text{.123kg}$

Concept used:

The velocity is

  $v=\sqrt{\frac{t}{m/l}}$    (1)

Calculation:

From equation (1),

  $\begin{array}{l}v=\sqrt{\frac{255\text{N}}{(0.123\text{kg/10m}}}\\ v=143.95\text{m/s}\end{array}$

Now,

  $\begin{array}{l}l=d_1+d_2\\ l=v{t}_1+v{t}_2\\ \text{or}\\ l=v{t}_1+v(t_2-2\times {10}^{-2})\end{array}$

Now,

  $\begin{array}{l}{t}_1=\frac{l+2\times {10}^{-2}v}{2v}\\ t_1=\frac{(10.0)+2\times {10}^{-2}(143.95)}{2(143.95)}\\ t_1=4.4726\times {10}^{-2}s\end{array}$

Now,

  $\begin{array}{c}d=v{t}_1\\ =(143.95\text{m/s)(4}\text{.4726}\times {10}^{-2}s)\\ =6.44\text{m}\end{array}$

Conclusion:

The distance where two pulses meet is $6.44\text{m}$ .