The upper end of a 3.80mlong steel wire is fastened to the ceiling, and a 54.0kg object is suspended from the lower end of the wire. You observe that it takes a transverse pulse 0.0492s to travel from the bottom to the top of the wire. What is the mass of the wire?

We know that the wave speed ofa string in terms of the tension T and the mass per unit lengthis given by,

$v=\sqrt{\frac{T}{\mu }}$

Here, linear mass density$\mu =\frac{m}{l}$

Newton’s 2^nd law: The acceleration of a body is directly proportional to the net force and inversely proportional to its mass.

Mathematically,$F=ma$

For hanged mass,$F=mg$ , where g is the value of gravity, i.e., 9.8m/s².

The length of the wire is l = 3.80m, the mass of the suspended object m = 54.0kg, the wave takes time t = 0.0492s to travel from the bottom to the top of the wire.

First, calculate the wave speed using the distance travelled and the time interval it takes:

$$\begin{gathered} v=\frac{l}{t} \\ =\frac{3.80m}{0.0492s} \\ =77.2m/s \end{gathered}$$

Apply Newton's second law to the hanged mass,

$$\begin{gathered} \sum _{}T=T-T_g \\ T=T_g=Mg \\ =54.0\times 9.8 \\ =529N \end{gathered}$$

Now put in the values for v and Tinto the wave speed on a string formula and solve for $\mu$,

$$\begin{gathered} 77.2=\sqrt{\frac{529}{\mu }} \\ \mu =\frac{529}{{\left(77.2\right)}^2} \\ =0.089kg/m \end{gathered}$$

Linear mass density is the mass per unit length, therefore modify the formula to find the mass of the wire,

$$\begin{gathered} \mu =\frac{m}{l} \\ m=\mu \times l \end{gathered}$$

Finally, put in the values for $\mu$and I into m expression,

$$\begin{gathered} m=\left(0.089kg/m\right)\times \left(3.80m\right) \\ =0.337kg \end{gathered}$$

Therefore, the mass of the wire is $m=0.337kg$.