91Ó°ÊÓ

Figures I_a, l_b, and l_c show the free-body diagrams. Depending on where the tension of the rope is considered, the tension in the rope assumes different values. Corresponding to each tension T there is a wave speed v given by the dynamic relationship

$v=\sqrt{\frac{F}{\mathrm{μ}}}$,the target variables in (a), (b) and (c).

In (d), compare the tension in the middle of the rope to the average of the tensions at the top and bottom of the rope as well as the wave speed at the middle of the rope to the average of the wave speeds at the top and bottom.

Due to the rope's weight, its tension is greater at the top F_top than at the middle F_middle than at the bottom F_bottom. Hence, the wave speed increases as a wave travels up the rope. V_top>v_middle>v_bottom, as in parts (a), (b), and (c). As of part (d), because the rope has a uniform mass, the tension in the rope increases linearly as the wave moves up along the rope, which explains why F_middle = F_avg. However, this is not true for the wave speed since the relationship between v and F is not linear.

Denote the total mass of the rope by$m_{rope}$.Expressing the rope's mass as$m_{rope}=\frac{w}{g}$,where w is the total weight of the rope, the rope's linear mass density is given by

$$\begin{gathered} \mathrm{μ}=\frac{m_{rope}}{L} \\ =\frac{w}{L_g} \end{gathered}$$

The tension role="math" localid="1664341025050" $F_{bottom}$ at the bottom of the rope is due to the hanging mass m. (Fig. l_a):

$F_{bottom}=mg$

Hence, the corresponding wave speed is

$$\begin{gathered} v_{bottom}=\sqrt{\frac{F_{bottom}}{\mathrm{μ}}} \\ =\sqrt{\frac{mg}{w/Lg}} \\ =g\sqrt{\frac{mL}{w}} \end{gathered}$$

Substitute the known values of L, w, and m:
$$\begin{gathered} v_{bottom}=\sqrt{\frac{F_{bottom}}{\mathrm{μ}}} \\ =\left(9.80m/s^2\right)\sqrt{\frac{\left(0.500kg\right)\left(6.00m\right)}{29.4N}} \\ =3.13m/s \end{gathered}$$

The tension$F_{middle}$ at the middle of the rope is due to the hanging mass m in addition to the lower half of the rope w/2 (Fig. 1_b):

$F_{middle}=mg+w/2$
Hence, the corresponding wave speed is
$$\begin{gathered} v_{middle}=\sqrt{\frac{F_{middle}}{\mathrm{μ}}} \\ =\sqrt{\frac{mg+w/2}{w/Lg}} \\ =\sqrt{Lg\left(\frac{mg}{w}+\frac{1}{2}\right)} \end{gathered}$$

Substitute the known values of L, w, and m:

$$\begin{gathered} v_{middle}=\sqrt{\frac{F_{middle}}{\mathrm{μ}}} \\ =\sqrt{\left(6.00m\right)\left(9.80m/s^2\right)\left(\frac{\left(0.500kg\right)\left(6.00m\right)}{29.4N}+\frac{1}{2}\right)} \\ =6.26m/s \end{gathered}$$

The tension$F_{top}$ at the top of the rope is due to the hanging mass m plus the total weight w of the rope (Fig. I_c):

$F_{top}=mg+w$

Hence, the corresponding wave speed

$$\begin{gathered} v_{top}=\sqrt{\frac{F_{top}}{\mathrm{μ}}} \\ =\sqrt{\frac{mg+w}{w/Lg}} \\ =\sqrt{Lg\left(\frac{mg}{w}+1\right)} \end{gathered}$$

Substituting the known values of L, w, and m, we find

$$\begin{gathered} v_{top}=\sqrt{\frac{F_{top}}{\mathrm{μ}}} \\ =\sqrt{\left(6.00m\right){\left(9.80m/s\right)}^2\left(\frac{\left(0.500kg\right)\left(6.00m\right)}{29.4N}+1\right)} \\ =8.28m/s \end{gathered}$$

The average of the tensions at the top and bottom of the rope is

$$\begin{gathered} F_{avg}=\frac{F_{top}+F_{bottom}}{2} \\ =\frac{mg+w+mg}{2} \\ =mg+\frac{w}{2} \end{gathered}$$

which is exactly the tension$F_{bottom}$ in the middle of the rope. However, the average of the wave speeds at the top

$$\begin{gathered} v_{avg}=\frac{v_{top}+v_{bottom}}{2} \\ =\frac{\left(8.28+3.13\right)m/s}{2} \\ =5.17m/s \end{gathered}$$

which is not equal to the wave speed at the middle of the rope $\left(v_{middle}=6.26m/s\right)$.

Therefore, the speed of the transverse waves at the (a) bottom = 3.13 m/s, (b) middle = 6.26 m/s, (c) top = 8.28 m/s. The tension in the middle of the rope is the average of the tensions at the top and bottom of the rope.The wave speed at the middle of the rope is not the average of the wave speeds at the top and bottom.