Chapter 18, Problem 70PQ

To determine

The relative minima of sound pressure located on the joining the two speakers.

Answer to Problem 70PQ

The relative minima of sound pressure located on the joining the two speakers are $1.80\text{m, }1.39\text{m, 0}\text{.990}\text{m, 0}\text{.586 m, 0}\text{.182 m, 2}\text{.20 m, 2}\text{.61 m, 3}\text{.01 m, 3}\text{.41 m and 3}\text{.82 m}\text{.}$

Explanation of Solution

Write an expression for the space between the nodes.

    $d_{nodes}=\frac{\lambda }{2}$                                                                                                                    (I)

Here, $d_{note}$ is the space between the nodes and $\lambda$ is the wavelength.

Write an expression for the wavelength.

    $\lambda =\frac{v}{f}$                                                                                                                      (II)

Here, $v$ is the wave speed and $f$ is the frequency.

Substitute equation (II) in figure (I).

    $\begin{array}{c}{d}_{nodes}=\frac{\left(\frac{v}{f}\right)}{2}\\ =\frac{v}{2f}\end{array}$                                                                                                          (III)

For speakers vibrating in phase, the antinode of the pressure is the point halfway between the speakers.

Write the expression for the distance of antinode of pressure from each speaker.

    $d=\frac{D}{2}$                                                                                                                   (IV)

Here, $d$ is the distance of antinode of pressure from each speaker and $D$ is the distance between the speakers.

There is a node one quarter of a wavelength away.

Write an expression for the distance of the node.

  $x=d-\frac{d_{nodes}}{2}$                                                                                                             (V)

Here, $x$ is the position of the first node.

After the node, there is a node one quarter of a wavelength.

Write an expression for the nodes.

  $x_{\min }=x-d_{nodes}$                                                                                                        (VI)

Here, $x_{\min }$ is the position of the node.

Figure 1 shows the position of minimum between the speaker.

Conclusion:

Substitute $343\text{m/s}$ for $v$ and $425{\text{s}}^{-1}$ for $f$ in equation (II) to find $d_{nodes}$.

    $\begin{array}{c}{d}_{nodes}=\frac{343\text{m/s}}{2\left(425{\text{s}}^{-1}\right)}\\ =\frac{343\text{m/s}}{850{\text{s}}^{-1}}\\ =0.404\text{m}\end{array}$

Substitute $4.00\text{m}$ for $D$ in equation (IV) to find $d$.

    $\begin{array}{c}d=\frac{4.00\text{m}}{2}\\ =2.00\text{m}\end{array}$

Substitute $2.00\text{m}$ for $d$ and $0.404\text{m}$ for $d_{nodes}$ in equation (V) to find $x$.

  $\begin{array}{c}x=2.00\text{m}-\frac{0.404\text{m}}{2}\\ =2.00\text{m}-0.202\text{m}\\ =1.80\text{m}\end{array}$

Substitute $1.80\text{m}$ for $x$ and $0.404\text{m}$ for $d_{nodes}$ in equation (VI) to find $x_{\min }$.

  $\begin{array}{c}{x}_{\min }=1.80\text{m}-0.404\text{m}\\ =1.39\text{m}\end{array}$

Substitute $1.39\text{m}$ for $x$ and $0.404\text{m}$ for $d_{nodes}$ in equation (VI) to find $x_{\min }$.

  $\begin{array}{c}{x}_{\min }=1.39\text{m}-0.404\text{m}\\ =0.990\text{m}\end{array}$

Substitute $0.990\text{m}$ for $x$ and $0.404\text{m}$ for $d_{nodes}$ in equation (VI) to find $x_{\min }$.

  $x_{\min }=0.586\text{m}$

Substitute $0.586\text{m}$ for $x$ and $0.404\text{m}$ for $d_{nodes}$ in equation (VI) to find $x_{\min }$.

  $\begin{array}{c}{x}_{\min }=0.586\text{m}-0.404\text{m}\\ =0.182\text{m}\end{array}$

Substitute $0.182\text{m}$ for $x$ and $-0.404\text{m}$ for $d_{nodes}$ in equation (VI) to find $x_{\min }$.

  $\begin{array}{c}{x}_{\min }=0.182\text{m}--0.404\text{m}\\ =2.20\text{m}\end{array}$

Substitute $2.20\text{m}$ for $x$ and $-0.404\text{m}$ for $d_{nodes}$ in equation (VI) to find $x_{\min }$.

  $\begin{array}{c}{x}_{\min }=2.20\text{m}--0.404\text{m}\\ =2.61\text{m}\end{array}$

Substitute $2.61\text{m}$ for $x$ and $-0.404\text{m}$ for $d_{nodes}$ in equation (VI) to find $x_{\min }$.

  $\begin{array}{c}{x}_{\min }=2.61\text{m}--0.404\text{m}\\ =3.01\text{m}\end{array}$

Substitute $3.01\text{m}$ for $x$ and $-0.404\text{m}$ for $d_{nodes}$ in equation (VI) to find $x_{\min }$.

  $\begin{array}{c}{x}_{\min }=3.01\text{m}--0.404\text{m}\\ =3.41\text{m}\end{array}$

Substitute $3.41\text{m}$ for $x$ and $-0.404\text{m}$ for $d_{nodes}$ in equation (VI) to find $x_{\min }$.

  $\begin{array}{c}{x}_{\min }=3.41\text{m}--0.404\text{m}\\ =3.82\text{m}\end{array}$

Thus, the relative minima of sound pressure located on the joining the two speakers are $1.80\text{m, }1.39\text{m, 0}\text{.990}\text{m, 0}\text{.586 m, 0}\text{.182 m, 2}\text{.20 m, 2}\text{.61 m, 3}\text{.01 m, 3}\text{.41 m and 3}\text{.82 m}\text{.}$