A train blows its horn as it approaches a crossing. A.) if the train engineer observes a frequency of 125Hz and a stationary observer at the crossing observes a frequency of 132Hz, what is the speed of the train? (The speed of sound is 330m/sec on this day). B.) what frequency is observed by the person at the crossing as the train moves away?
Divide velocity by wavelength for frequency/ speed??? #8179081423 text if available when done please.
http://hyperphysics.phy-astr.gsu.edu/hbase/Sound/dopp.html
To find the speed of the train, we can use the Doppler effect. The Doppler effect is the change in frequency or wavelength of a wave observed by an observer moving relative to the source of the waves. In this case, the observer at the crossing is stationary and observing the train's horn.
A.) Let's first calculate the speed of the train. The equation used in this scenario is:
(v + vo)/(v + vs) = f/f',
where:
- v is the velocity of sound (330 m/s),
- vo is the observer's velocity (0 m/s since the observer is stationary),
- vs is the source's velocity (train's velocity),
- f is the source's frequency (125 Hz),
- f' is the observer's observed frequency (132 Hz).
Now, rearranging the equation to solve for vs, the train's velocity, we get:
(vs - vo) = (f/f') * v - v,
vs = [(f/f') * v - v] + vo.
Plugging in the given values:
vs = [(125 Hz / 132 Hz) * 330 m/s - 330 m/s] + 0 m/s.
Calculating further:
vs = (125/132 * 330 - 330) m/s,
vs = 20 m/s.
Therefore, the speed of the train is 20 m/s.
B.) Now, let's calculate the frequency observed by the person at the crossing as the train moves away. In this case, the source (train) is moving away from the observer.
The equation used for this scenario is:
(v - vo)/(v - vs) = f/f',
where:
- vo is still 0 m/s,
- vs is still the train's velocity (20 m/s),
- f is now the new source's frequency,
- f' is the observer's observed frequency (which we need to find).
Rearranging the equation to solve for f', the observer's observed frequency, we get:
(f' - f) = (f/v) * (vs - vo),
f' = f - (f/v) * (vs - vo).
Plugging in the given values:
f' = 125 Hz - (125 Hz / 330 m/s) * (20 m/s - 0 m/s).
Calculating further:
f' = 125 Hz - (125 Hz / 330 m/s) * 20 m/s,
f' = 125 Hz - 75.758 Hz,
f' ≈ 49.242 Hz.
Therefore, the frequency observed by the person at the crossing as the train moves away is approximately 49.242 Hz.