Q62P - University Physics with Modern Physics

Question

A uniform 165-Nbar is supported horizontally by two identical wires A and B (Fig. P16.62). A small 185-N cube of lead is placed three fourths of the way from A to B. The wires are each 75.0 cm long and have a mass of 5.50 g. If both of them are simultaneously plucked at the centre, what is the frequency of the beats that they will produce when vibrating in their fundamental?

                                   

Step-by-Step Solution

Answer

The beat frequency produces by the two wires is 27.46 Hz.

STEP 1 Concept on net torque and frequency of wire

$\frac{\mathrm{x}}{2}\left({\mathrm{W}}_{\mathrm{b}}\right)+\frac{3\mathrm{x}}{4}\left({\mathrm{W}}_{\mathrm{C}}\right)-\mathrm{x}\left({\mathrm{F}}_{\mathrm{b}}\right)=0$ where, x is the length of the bar. ${\mathrm{W}}_{\mathrm{b}}$  is the weight of the bar.   is the weight of the cube. The frequency of wire is were $\mathrm{f}=\frac{1}{2\mathrm{L}}\sqrt{\frac{\mathrm{F}}{{\displaystyle \frac{\mathrm{m}}{\mathrm{L}}}}}$ is the fundamental frequency of the wire, L is the length of the wire, F is the tension in the wire , m is the mass of the wire

STEP 2 Calculate the tension and frequency on the wire B

The weight of the bar is 165 N, the bar is supported by two identical wire at its both the end, a small 185 N cube of lead is placed three-fourth of the way from A to B, the length of the each wire is 75.0 cm and the mass of the each wire is 5.50g.

At equilibrium, the net torque about the point A is zero. Substitute the values in equation $\frac{\mathrm{x}}{2}\left({\mathrm{W}}_{\mathrm{b}}\right)+\frac{3\mathrm{x}}{4}\left({\mathrm{W}}_{\mathrm{C}}\right)-\mathrm{x}\left({\mathrm{F}}_{\mathrm{b}}\right)=0$ to find ${\mathrm{F}}_{\mathrm{b}}$

$$\begin{gathered} \frac{\mathrm{x}}{2}\left(165\mathrm{N}\right)+\frac{3\mathrm{x}}{4}\left(185\mathrm{N}\right)-\mathrm{x}\left({\mathrm{F}}_{\mathrm{b}}\right)=0 \\ {\mathrm{F}}_{\mathrm{B}-}221.25\mathrm{N} \end{gathered}$$

Therefore, the tension on the wire B is 221.25N

Substitute the values in equation ${\mathrm{f}}_{\mathrm{B}}=\frac{1}{2\mathrm{L}}\sqrt{\frac{{\mathrm{F}}_{\mathrm{B}}}{{\displaystyle \frac{\mathrm{m}}{\mathrm{L}}}}}$ to find ${\mathrm{f}}_{\mathrm{b}}$

 $$\begin{gathered} {\mathrm{f}}_{\mathrm{B}}=\frac{1}{2\left(75.0\times {\displaystyle \frac{1}{100}}\right)}\sqrt{\frac{221.25}{{\displaystyle \frac{\mathrm{m}}{\left(75.0\times {\displaystyle \frac{1}{100}}\right)}}}} \\ =115.79\mathrm{Hz} \end{gathered}$$

Thus, the fundamental frequency of the wire B is 115.79 Hz.

STEP 3 Calculate the frequency on the wire of the wire A

At equilibrium, the net force along the y axis is zero. ${\mathrm{F}}_{\mathrm{A}}+{\mathrm{F}}_{\mathrm{B}}-{\mathrm{w}}_{\mathrm{b}}-{\mathrm{w}}_{\mathrm{c}}=0$

Substitute 221.25 N for 165 N for ${\mathrm{w}}_{\mathrm{b}}$ and 185 N for ${\mathrm{w}}_{\mathrm{c}}$ in the above equation to find ${\mathrm{F}}_{\mathrm{A}}$

$$\begin{gathered} {\mathrm{F}}_{\mathrm{A}}+221.25-165-185=0 \\ {\mathrm{F}}_{\mathrm{A}}=128.75\mathrm{N} \end{gathered}$$

The fundamental frequency of the wire A is. ${\mathrm{f}}_{\mathrm{A}}=\frac{1}{2\mathrm{L}}\sqrt{\frac{{\mathrm{F}}_{\mathrm{A}}}{{\displaystyle \frac{\mathrm{m}}{\mathrm{L}}}}}$ Substitute 128.75 N for ${\mathrm{f}}_{\mathrm{A}}$ , 75.0 cm for L and 5.50g  for m in the above equation to find ${\mathrm{f}}_{\mathrm{A}}$

$$\begin{gathered} {\mathrm{f}}_{\mathrm{A}}=\frac{1}{2\left(75.0\times {\displaystyle \frac{1}{100}}\right)}\sqrt{\frac{128.75}{{\displaystyle \frac{\mathrm{m}}{\left(75.0\times {\displaystyle \frac{1}{100}}\right)}}}} \\ =88.33\mathrm{Hz} \end{gathered}$$

STEP 4 Calculate the beat frequency .

The beat frequency produces by the two wires is ${\mathrm{f}}_{\mathrm{beat}}={\mathrm{f}}_{\mathrm{B}}-{\mathrm{f}}_{\mathrm{A}}$ Substitute 115.79 Hz for ${\mathrm{f}}_{\mathrm{B}}$ and 88.33 Hz for ${\mathrm{f}}_{\mathrm{A}}$ in the above equation to find ${\mathrm{f}}_{\mathrm{beat}}$

$$\begin{gathered} {\mathrm{f}}_{\mathrm{beat}}=115.79\mathrm{Hz}-88.33\mathrm{Hz} \\ =27.89\mathrm{Hz} \end{gathered}$$

Therefore, the beat frequency produces by the two wires is 27.46 Hz.