Question

A trumpet player on a moving railroad flatcar moves toward a second trumpet player standing alongside the track while both play a $\mathbf{440}\mathbf{\text{\hspace{0.17em}Hz}}$note. The sound waves heard by a stationary observer between the two players have a beat frequency of$\mathbf{4.0}\mathbf{\text{\hspace{0.17em}beats}}\mathbf{/}\mathbf{\text{s}}$.What is the flatcar’s speed?

Short answer

The flatcar’s speed is, $3.1\text{\hspace{0.17em}m}/\text{s}$.

Step-by-step solution

The given data

1. The frequency of trumpet player is $\mathrm{f}=440\text{\hspace{0.17em}Hz}$.
2. The beat frequency is$\mathrm{\Delta f}=4.0\text{\hspace{0.17em}beats}/\text{s}$ .
3. The speed of sound is, $\mathrm{v}=343\text{\hspace{0.17em}m}/\text{s}$.

Understanding the concept of Doppler’s Effect 

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)

$\frac{\mathbf{f}\mathbf{\text{'}}}{\mathbf{f}}\mathbf{=}\frac{\mathbf{v}}{\mathbf{v}\mathbf{-}{\mathbf{v}}_{\mathbf{s}}}$ …(i)

Calculation of speed of the flatcar

We are given that the beat frequency as:

$\begin{array}{c}{\mathrm{f}}^{\text{'}}-\mathrm{f}=4\text{\hspace{0.17em}Hz}\\ {\mathrm{f}}^{\text{'}}=\mathrm{f}+4\text{\hspace{0.17em}Hz}\end{array}$

Substitute the value of f in the above equation.

$\begin{array}{c}{\mathrm{f}}^{\text{'}}=\mathrm{f}+4\text{\hspace{0.17em}Hz}\\ =440\text{\hspace{0.17em}Hz}+4\text{\hspace{0.17em}Hz}\\ {\mathrm{f}}^{\text{'}}=444\text{\hspace{0.17em}Hz}\end{array}$

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

$\begin{array}{c}\frac{444\text{\hspace{0.17em}Hz}}{440\text{\hspace{0.17em}Hz}}=\frac{343\text{\hspace{0.17em}m}/\text{s}}{343\text{\hspace{0.17em}m}/\text{s}-{\mathrm{v}}_{\mathrm{s}}}\\ 1.007=\frac{343\text{\hspace{0.17em}m}/\text{s}}{343\text{\hspace{0.17em}m}/\text{s}-{\mathrm{V}}_{\mathrm{s}}}\\ (343\text{\hspace{0.17em}m}/\text{s}-{\mathrm{v}}_{\mathrm{s}})=\frac{343\text{\hspace{0.17em}m}/\text{s}}{1.009}\\ 343\text{\hspace{0.17em}m}/\text{s}-\frac{343\text{\hspace{0.17em}m}/\text{s}}{1.009}={\mathrm{v}}_{\mathrm{s}}\\ {\mathrm{v}}_{\mathrm{s}}=3.059\text{\hspace{0.17em}m}/\text{s}\\ {\mathrm{v}}_{\mathrm{s}}\approx 3.1\text{\hspace{0.17em}m}/\text{s}\end{array}$

Hence, the value of the speed of flatcar is$3.1\text{\hspace{0.17em}m}/\text{s}$ .