Question

FIGURE P16.58 shows two masses hanging from a steel wire.

The mass of the wire is 60.0 g. A wave pulse travels along the wire from point 1 to point 2 in 24.0 ms. What is mass m?

Short answer

The mass of each of the block is,$17.8kg.$

Step-by-step solution

Introduction

Two masses have a distance of, $d=4m$

Time of pulse, $t=24ms=24\times \times {10}^{-3}s$

Total mass of the wire, $M=60g$

Determination of tension

The pulse has a speed of,

$$\begin{gathered} v=dt \\ v=\frac{4.0m}{24.0\times {10}^{-3}s} \\ v=166.7m/s \end{gathered}$$

Now, mass of middle part of the wire is,

$$\begin{gathered} m=\frac{M}{2} \\ m=\frac{60g}{2} \\ m=30g \\ m=0.03kg \end{gathered}$$

Therefore, the tension in the middle portion of the wire can be obtained by,

$$\begin{gathered} \sqrt{\frac{T}{\mu }}=v \\ T=v^2\mu \\ T=v^2\left(\frac{m}{L}\right) \\ T={(166.7m/s)}^2\left(\frac{0.03kg}{4.0m}\right)T=208.4N \end{gathered}$$

Consider vertical and horizontal component of wire as $T_y,T_x$

The horizontal component of tension$T_x=T=208.4N$

Therefore,

$$\begin{gathered} \frac{T_y}{T_x}=\tan {40}^o \\ \frac{T_y}{T}=\tan {40}^o \\ {T}_y=(208.4N)\tan \left({40}^o\right) \\ {T}_y=174.9N \end{gathered}$$

Determination of mass

Now, vertical component of the tension $T_y$= block weightrole="math" localid="1648839050885" $\left(m_{b}g\right)$

Therefore, mass of each block,

$$\begin{gathered} m_{b}g=T_y \\ {m}_b=\frac{T_y}{g} \\ {m}_b=\frac{174.9N}{9.80m/s^2} \\ {m}_b=17.8kg \end{gathered}$$