91Ó°ÊÓ

1. The frequency of trumpet player is $\mathrm{f}=440\text{ H³ú}$.
2. The beat frequency is$\mathrm{Δ´Ú}=4.0\text{ b±ð²¹³Ù²õ}/\text{s}$ .
3. The speed of sound is, $\mathrm{v}=343\text{″¾}/\text{s}$.

We are given beats and frequency. From that, we can findthefrequency ratio. Using Doppler’s formula, we can find the velocity oftheflatcar.

Formula:

The frequency of the sound wave according to the Doppler Effect, (where, the velocity of the observer is 0)

$\frac{\mathbf{f}\mathbf{\text{'}}}{\mathbf{f}}\mathbf{=}\frac{\mathbf{v}}{\mathbf{v}\mathbf{−}{\mathbf{v}}_{\mathbf{s}}}$ …(¾±)

We are given that the beat frequency as:

$\begin{array}{c}{\mathrm{f}}^{\text{'}}−\mathrm{f}=4\text{ H³ú}\\ {\mathrm{f}}^{\text{'}}=\mathrm{f}+4\text{ H³ú}\end{array}$

Substitute the value of f in the above equation.

$\begin{array}{c}{\mathrm{f}}^{\text{'}}=\mathrm{f}+4\text{ H³ú}\\ =440\text{ H³ú}+4\text{ H³ú}\\ {\mathrm{f}}^{\text{'}}=444\text{ H³ú}\end{array}$

Now we have to use the following formula of equation (i), we can find the speed of the flatcar as given:

$\begin{array}{c}\frac{444\text{ H³ú}}{440\text{ H³ú}}=\frac{343\text{″¾}/\text{s}}{343\text{″¾}/\text{s}−{\mathrm{v}}_{\mathrm{s}}}\\ 1.007=\frac{343\text{″¾}/\text{s}}{343\text{″¾}/\text{s}−{\mathrm{V}}_{\mathrm{s}}}\\ (343\text{″¾}/\text{s}−{\mathrm{v}}_{\mathrm{s}})=\frac{343\text{″¾}/\text{s}}{1.009}\\ 343\text{″¾}/\text{s}−\frac{343\text{″¾}/\text{s}}{1.009}={\mathrm{v}}_{\mathrm{s}}\\ {\mathrm{v}}_{\mathrm{s}}=3.059\text{″¾}/\text{s}\\ {\mathrm{v}}_{\mathrm{s}}≈3.1\text{″¾}/\text{s}\end{array}$

Hence, the value of the speed of flatcar is$3.1\text{″¾}/\text{s}$ .