91Ó°ÊÓ

• 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,

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

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,

$\mathrm{Δ}t=\frac{d}{v}$……. (i)

In the figure, in the right angle triangle,

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

Using these values in equation 1,

$\mathrm{Δ}t=\frac{D\sin \mathrm{θ}}{v}$…… (ii)

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

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$\mathrm{θ}={90}^0,   \sin 90=1.$

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

$\mathrm{Δ}t=\frac{D}{v_w}$

In the figure, in the right angle triangle,

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

Therefore,

$\mathrm{Δ}t=\frac{D}{1482}$

Time delay in water to reach from left ear to right ear is,
$\mathrm{Δ}t=\frac{D}{v_w}$ …â¶Ä¦(¾±¾±¾±)

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

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

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

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

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

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