Q48E - University Physics with Modern Physics

Question

(a) A sound source producing 1.00-kHz waves moves toward a stationary listener at one-half the speed of sound. What frequency will the listener hear? (b) Suppose instead that the source is stationary and the listener moves toward the source at one-half the speed of sound. What frequency does the listener hear? How does your answer compare to that in part (a)? Explain on physical grounds why the two answers differ.

Step-by-Step Solution

Answer

(a) when the source moves towards the listener the frequency heard by him would be 2000Hz

(b) when the listener moves towards the source the frequency heard by him would be 1500Hz.

Step1: Dopplers equation

$f_L=\left(\frac{v+v_L}{v+v_s}\right)f_s$ where ${\mathrm{f}}_{\mathrm{L}}$is the observer frequency of sound, v is the speed of sound waves, $V_{\mathrm{S}}$ is observer velocity, ${\mathrm{V}}_{\mathrm{L}}$ is source velocity and  ${\mathrm{f}}_{\mathrm{s}}$ is actual frequency of sound waves.

Step2: Calculation

${\mathrm{f}}_{\mathrm{s}}=1000\mathrm{Hz}$, The positive direction is from the listener to the source $\mathrm{v}=344\mathrm{m}/\mathrm{s}$.

(a)

$$\begin{gathered} \mathrm{v}=\left(344\mathrm{m}/\mathrm{s}\right) /2 \\ =-172 \mathrm{m}/\mathrm{s} \\ {\mathrm{v}}_{\mathrm{L}}=0\mathrm{m}/\mathrm{s} \end{gathered}$$

Apply in doppler’s equation,

$$\begin{gathered} {\mathrm{f}}_{\mathrm{L}}=\left(\frac{\mathrm{v}+{\mathrm{v}}_{\mathrm{L}}}{\mathrm{v}+{\mathrm{v}}_{\mathrm{s}}}\right){\mathrm{f}}_{\mathrm{s}} \\ =\left(\frac{344\mathrm{m}/ \mathrm{s}}{344\mathrm{m} /\mathrm{s}-172\mathrm{m} /\mathrm{s}}\right)\left(1000\mathrm{Hz}\right) \\ =2000\mathrm{Hz} \end{gathered}$$

Hence, when the source moves towards the listener the frequency heard by him would be 2000Hz.

(b) ${\mathrm{v}}_{\mathrm{s}}=0,{\mathrm{v}}_{\mathrm{L}}=172\mathrm{m}/\mathrm{s}$

Apply in doppler’s equation,

$$\begin{gathered} {\mathrm{f}}_{\mathrm{L}}=\left(\frac{\mathrm{v}+{\mathrm{v}}_{\mathrm{L}}}{\mathrm{v}+{\mathrm{v}}_{\mathrm{s}}}\right){\mathrm{f}}_{\mathrm{s}} \\ =\left(\frac{344\mathrm{m}/ \mathrm{s}+172\mathrm{m} /\mathrm{s}}{344\mathrm{m} /\mathrm{s}}\right)\left(1000\mathrm{Hz}\right) \\ =1500\mathrm{Hz} \end{gathered}$$

Hence, when the listener moves towards the source the frequency heard by him would be 1500Hz.

The answer in (b) is much less than the answer in (a). It is the velocity of the source and listener relative to the air that determines the effect, not the relative velocity of the source and listener relative to each other.