Question

Straight line$\mathit{A}\mathit{B}$connects two point sources that are apart, emit$\mathbf{300}\mathbf{Hz}$ sound waves of the same amplitude, and emit exactly out of phase. (a) What is the shortest distance between the midpoint of $\mathit{A}\mathit{B}$and a point onwhere the interfering waves cause maximum oscillation of the air molecules? What are the (b) second and (c) third shortest distances?

Short answer

1. The shortest distance between the midpoint of AB and point on AB where the interfering waves cause maximum oscillation of the air molecule is 0
2. Second shortest distance between the midpoint of AB and point on AB where the interfering waves cause maximum oscillation of the air molecule is$0.572\text{\hspace{0.17em}m}$ .
3. Third shortest distance between the midpoint of AB and point on AB where the interfering waves cause maximum oscillation of the air molecule is $1.14\text{\hspace{0.17em}m}$.

Step-by-step solution

The given data

1. Distance between AB is $5.0\text{\hspace{0.17em}m}$.
2. Frequency of sources,$f=300\text{\hspace{0.17em}Hz}$.
3. Velocity of sound,$v=343\text{\hspace{0.17em}m}/\text{s}$
4. Amplitude of waves is the same, and they are out of phase.

Understanding the concept of nodes and interference

We can write the distance between two nodes in terms of the wavelength of sound waves. Then we can write it in terms of velocity and frequency using the corresponding relation. Then applying the condition for constructive interference we can find the shortest, second shortest, and the third shortest distance between the midpoint of AB and the point on AB where the interfering waves cause maximum oscillation of the air molecule.

Formula:

The wavelength of a body in wave motion,

$\mathit{\lambda }\mathbf{=}\frac{\mathbf{v}}{\mathbf{f}}$ …(i)

The distance between two nodes of the wave,

$\mathit{\Delta }\mathit{x}\mathbf{=}\frac{\mathbf{\lambda }}{\mathbf{2}}$ …(ii)

The position value of nodes,

$\mathit{x}\mathbf{=}\mathit{n}\frac{\mathbf{\lambda }}{\mathbf{2}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{\text{Where}}\mathbf{,}\mathit{n}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{,}\mathbf{1}\mathbf{,}\mathbf{2}\mathbf{,}\mathbf{3}$ …(iii)

a) Calculation of the shortest wave

As it is given in the problem, it is clear that the waves are out of phase and that will cancel at a point exactly midway between them.

Distance between two nodes is given by substituting equation (i) in equation (ii) as:

$\Delta x=\frac{v}{2f}$

Substitute all the value in the above equation.

$\begin{array}{c}\Delta x=\frac{343\text{\hspace{0.17em}m}/\text{s}}{2\times \left(300\text{\hspace{0.17em}Hz}\right)}\\ =0.5716\text{\hspace{0.17em}m}\\ \Delta x\simeq 0.572\text{\hspace{0.17em}m}\end{array}$

Now, nodes are found at distance using equation (ii) and (iii),

$x=n\Delta x$

Substitute all the value in the above equation.

$x=n\left(0.572\text{\hspace{0.17em}m}\right)$ …(a)

Where, $n=0,\pm 1,\pm 2\dots \mathrm{..}$

Hence, the first shortest distance from the midpoint where nodes are found is, $0\text{\hspace{0.17em}m}$.

b) Calculation of second-shortest distance

The second shortest distance from the midpoint where nodes are found using equation (a) is given as:

For, $n=1$

$\begin{array}{c}x=1\times \left(0.572\text{\hspace{0.17em}m}\right)\\ x=0.572\text{\hspace{0.17em}m}\end{array}$

Hence, the second shortest distance is $0.572m$

c) Calculation of third shortest distance

The third shortest distance from the midpoint where nodes are found using equation (a) is given as:

For, $n=2$

$\begin{array}{c}x=2\times \left(0.572\text{\hspace{0.17em}m}\right)\\ x=1.144\text{\hspace{0.17em}m}\end{array}$

Hence, the value of third shortest distance is $1.14\text{\hspace{0.17em}m}$.