91Ó°ÊÓ

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

$v=\sqrt{\frac{T}{\mathrm{μ}}}$

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

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,

$$\begin{gathered} \underset{}{∑}T=T-T_g \\ T=T_g=Mg \\ =54.0×9.8 \\ =529N \end{gathered}$$

Now put in the values for v and Tinto the wave speed on a string formula and solve for $\mathrm{μ}$,

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

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

$$\begin{gathered} \mathrm{μ}=\frac{m}{l} \\ m=\mathrm{μ}×l \end{gathered}$$

Finally, put in the values for $\mathrm{μ}$and I into m expression,

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

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