Q17E - University Physics with Modern Physics

Question

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?

Step-by-Step Solution

Answer

The mass of the wire is 0.337kg.

Step 1:Determination of the formula for Mechanical Waves

We know that the wave speed ofa string in terms of the tension T and the mass per unit length is 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.

Step 2: Calculation for the mass of the wire by using wave speed formula

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,

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

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