17.7 The doppler effect  (Page 4/8)

If the observer is moving away from the source ( [link] ), the observed frequency can be found:

The equations for an observer moving toward or away from a stationary source can be combined into one equation:

where $f_{\text{o}}$ is the observed frequency, $f_{\text{s}}$ is the source frequency, $v_{\text{w}}$ is the speed of sound, $v_{\text{o}}$ is the speed of the observer, the top sign is for the observer approaching the source and the bottom sign is for the observer departing from the source.

[link] and [link] can be summarized in one equation (the top sign is for approaching) and is further illustrated in [link] :

where $f_o$ is the observed frequency, $f_s$ is the source frequency, $v_w$ is the speed of sound, $v_o$ is the speed of the observer, $v_s$ is the speed of the source, the top sign is for approaching and the bottom sign is for departing.

Calculating a doppler shift

(a) What frequencies are observed by a stationary person at the side of the tracks as the train approaches and after it passes?

(b) What frequency is observed by the train’s engineer traveling on the train?

Strategy

Solution

1. Enter known values into $f_{\text{o}}=f_{\text{s}}\left(\frac{v}{v-v_{\text{s}}}\right):$
   
   $f_{\text{o}}=f_{\text{s}}\left(\frac{v}{v-v_{\text{s}}}\right)=\left(150\text{Hz}\right)\left(\frac{340\text{m/s}}{340\text{m/s}-35.0\text{m/s}}\right).$
   
   Calculate the frequency observed by a stationary person as the train approaches:
   
   $f_{\text{o}}=\left(150\text{Hz}\right)\left(1.11\right)=167\text{Hz}\text{.}$
   
   Use the same equation with the plus sign to find the frequency heard by a stationary person as the train recedes:
   
   $f_{\text{o}}=f_{\text{s}}\left(\frac{v}{v+v_{\text{s}}}\right)=\left(150\text{Hz}\right)\left(\frac{340\text{m/s}}{340\text{m/s}+35.0\text{m/s}}\right).$
   
   Calculate the second frequency:
   
   $f_{\text{o}}=\left(150\text{Hz}\right)\left(0.907\right)=136\text{Hz}\text{.}$
2. Identify knowns:
   • It seems reasonable that the engineer would receive the same frequency as emitted by the horn, because the relative velocity between them is zero.
   • Relative to the medium (air), the speeds are $v_{\text{s}}=v_{\text{o}}=35.0\text{m/s}\text{.}$
   • The first Doppler shift is for the moving observer; the second is for the moving source.
   
   Use the following equation:
   
   $f_{\text{o}}=\left[f_{\text{s}}\left(\frac{v\pm v_{\text{o}}}{v}\right)\right]\left(\frac{v}{v\mp v_{\text{s}}}\right).$
   
   The quantity in the square brackets is the Doppler-shifted frequency due to a moving observer. The factor on the right is the effect of the moving source.
   Because the train engineer is moving in the direction toward the horn, we must use the plus sign for $v_{\text{obs}};$ however, because the horn is also moving in the direction away from the engineer, we also use the plus sign for $v_{\text{s}}.$ But the train is carrying both the engineer and the horn at the same velocity, so $v_{\text{s}}=v_{\text{o}}.$ As a result, everything but $f_{\text{s}}$ cancels, yielding
   
   $f_{\text{o}}=f_{\text{s}}.$