Doppler!
A boy is standing in front of a stationary train. The train blows a horn of frequency 400 Hz. If the wind is blowing from the train to the boy at a speed of 30 m/s, what will be the frequency of the sound the boy hears?
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4 solutions
Good solution . What will be your answer to this: A train is moving towards a platform with the speed of 1 5 m / s and the wind is blowing on the direction opposite to the direction of train's motion , what is the observed frequency of the pasanger who is standing on the platform? Original frequency of horn f o = 3 0 0 H z and v a i r = 3 0 m / s , v s o u n d = 3 3 0 m / s
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υ / = ( 3 3 0 − 3 0 − 1 5 3 3 0 − 3 0 − 0 ) 3 0 0 = 3 1 5 . 7 H z
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hmm. The formula is good . Well done. :)
Hello! I think we can still use the classic formula : ν ′ = s − v s s − v l If we consider that v l and v s are respectively the speed of the listener and the speed of the source RELATIVE TO THE MEDIUM, which therefore takes the speed of the wind into account.
Use Basic Definition of Doppler's effect :: "Since There is No Relative Motion Between Source and Observe So apparent frequency will be equal to frequency of Source "
. What will be your answer to this: A train is moving towards a platform with the speed of 1 5 m / s and the wind is blowing on the direction opposite to the direction of train's motion , what is the observed frequency of the pasanger who is standing on the platform? Original frequency of horn f o = 3 0 0 H z and v a i r = 3 0 m / s , v s o u n d = 3 3 0 m / s
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3 1 5 . 7 H z
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That's right. Why do we take the wind speed into consideration in this case?
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@Keshav Tiwari – In the former case, even if you take wind speed in account it won't have the net effect in the Source-Listener system, whereas, on the other hand, in the latter case, since the source is moving, the wind speed gets a cumulative effect due to the change in speed of source.
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@A Former Brilliant Member – thanks ! Got it ! :)
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@Keshav Tiwari – Anytime...and thanks for following me!!:):)
Although sound is a wave of compressions and rarefactions, let's talk of a wave 'crest' as a shorthand for maximum compression. The boy and train remain in position and every crest emitted by the train is received sometime later by the boy.
Since there is no change in the conditions from one moment to the next, we can imagine the boy and the train continuing for some time, and so (because crests are neither created nor destroyed between the train and the boy) if the train emits 400 crest per second the boy must receive 400 crests per second.
Due to wind the speed of sound will be affected. If the wind is blowing along the sound wave it's speed increases and vice versa. In this case the wind is blowing along the sound wave.So speed of sound increases. It will be equal to 370m/s(340+30). Substituting this in the formula we get the answer as 400Hz.
Sure we can apply the Doppler shift formula, but what better way than to see for yourself how the waves actually propagate.
Assuming the horn makes physical disturbances are a rate of 400Hz, then there is a crest once every 2.5ms. The first crest leaves the horn and travels in the air at a speed of sound + speed of the wind, which is 340m/s + 30m/s, which gets the the boy "sooner" than without the wind. However, the next crest comes at 2.5ms later and still travels at 370m/s towards the boy. The boy's eardrums detect this 2.5ms later. Repeat, and the boy's eardrums will be pounded at a rate of one push per 2.5ms or a frequency of 400Hz.
If you want to use the usual Doppler Shift formula, transform into a reference frame where there is no wind, and you will find that the boy and the train are moving in the same direction.
In this case, the conventional arrow that helps you figure out the signs of the relative motions is drawn from the listener to the source and their velocities relative to the air (medium) is +30m/s.
According to the Doppler Shift equation, we have f S f L = v + v s v + v L = 3 4 0 + 3 0 3 4 0 + 3 0 = 1
Thus, the frequency does not shift.
Apparent frequency, υ / = ( s + w − v s s + w − v l ) υ , w h e r e υ ⟶ o r i g i n a l f r e q u e n c y s ⟶ s p e e d o f s o u n d i n m e d i u m w ⟶ s p e e d o f w i n d v l ⟶ v e l o c i t y o f l i s t n e r v s ⟶ v e l o c t y o f s o u r c e
Now substituting values,
υ / = ( 3 4 0 + 3 0 − 0 3 4 0 + 3 0 − 0 ) 4 0 0 = 4 0 0 ㎐
[Note:this formula can be used for other questions on Doppler effect but need take care of sign convention]