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?

$\frac{\mathrm{x}}{2}\left({\mathrm{W}}_{\mathrm{b}}\right)+\frac{3\mathrm{x}}{4}\left({\mathrm{W}}_{\mathrm{C}}\right)-\mathrm{x}\left({\mathrm{F}}_{\mathrm{b}}\right)=0$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

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.

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${\mathrm{f}}_{\mathrm{A}}$

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

The beat frequency produces by the two wires is${\mathrm{f}}_{\mathrm{beat}}={\mathrm{f}}_{\mathrm{B}}-{\mathrm{f}}_{\mathrm{A}}$ Substitute115.79 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.