Question:One clue used by your brain to determine the direction of a source of sound is the delay$∆t$between the arrival of the sound at the ear closer to the source and the arrival at the father ear. Assume that the source is distant so that a wave front from it is approximately planar when it reaches you , and let Drepresent the separation between your ears.(a) If the source is located at anglein front of you figure , what is$∆t$in term ofD and the speed of soundin air? (b) If you are submerged in water and the speed of sound is directly to your right, what is$∆t$in terms of V_wand the speed of soundin water? (c) Based on the time –delay clue, your brain interprets the submerged sound to arrive at an angle$\theta$ from the forward direction. Evaluate$\theta$for fresh water at 20 ⁰C.

• When the wave front reaches the ear, it is planar.
• Time delay between the arrival of the sound for two ear is$∆t$.
• Separation between two ears is D.
• Source is located at an angle θ.

By using the distance time relation and trigonometric relation in the given diagram, find time in air and in water. Taking the ratio of ${\mathit{v}}_{\mathbf{a}\mathbf{i}\mathbf{r}\mathbf{}\mathbf{}}\mathit{a}\mathit{n}\mathit{d}\mathbf{}{\mathit{v}}_{\mathbf{w}\mathbf{a}\mathbf{t}\mathbf{e}\mathbf{r}}$, find angle θ.

The expression for the delay time due to distance is given by,

$\mathit{\Delta }\mathit{t}\mathbf{=}\frac{\mathbf{d}}{\mathbf{v}}$

Here v is the speed of the sound at 20⁰C , and the value of v is 343 m/s.

Time delay is due to the distance d that the wave front must travel to reach the left ear and then right ear. Now, if is the velocity of sound in air, then time delay can be calculated as,

$\Delta t=\frac{d}{v}$……. (i)

In the figure, in the right angle triangle,

role="math" localid="1661340341270" $$\begin{gathered} \sin \theta =\frac{d}{D} \\ d=D\sin \theta \end{gathered}$$

Using these values in equation 1,

$\Delta t=\frac{D\sin \theta }{v}$…… (ii)

Hence,$\Delta t$ in term of D and the speed of sound v in air is$\Delta t=\frac{D\sin \theta }{v}$ .

Now, as the medium is water, if the velocity of the sound in water is V_w.

And the source is directly towards the right ear, therefore$\theta ={90}^0, \sin 90=1.$

Hence, the time delay in water can be calculated as,

$\Delta t=\frac{D}{v_w}$

In the figure, in the right angle triangle,

$$\begin{gathered} \sin \theta =\frac{d}{D} \\ d=D\sin \theta \\ =D\sin 90° \\ =D \end{gathered}$$

Therefore,

$\Delta t=\frac{D}{1482}$

Time delay in water to reach from left ear to right ear is,
$\Delta t=\frac{D}{v_w}$ ……(iii)

Hence,$\Delta t$ in term of D and the speed of sound v in water is $\Delta t=\frac{D}{v_w}$.

To find angle equating equation (ii) and (iii),

$$\begin{gathered} \frac{D\sin \theta }{v}=\frac{D}{v_w} \\ \sin \theta =\frac{v}{v_w} \end{gathered}$$

Substitute 343 m/s for v and 1482 m/s for V_w into the above equation,

$$\begin{gathered} \sin \theta =\frac{343}{1482} \\ =0.231 \\ \theta ={\sin }^{-1}\left(0.231\right) \\ ={13}^0 \end{gathered}$$

Hence, the $\theta ={13}^0$for fresh water at 20⁰C is $\theta ={13}^0$.