Ivan M.
asked • 03/30/22Pitch Problems Help! (Frequencies and Sound)
Two trumpet players are riding in separate convertibles which are moving in opposite directions at a speed of 30.0 m/sec. They both strike a note with a frequency of 1024 Hz. Calculate:
A.) The pitch heard coming from one vehicle by a listener of the other vehicle.
B.) Tbe pitch heard coming from either vehicle by an observer stationed directly between both vehicles.
C.) The pity heard by a listener in either vehicle if both vehicles turn around and move toward each other at the same speeds.
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1 Expert Answer
This is a Doppler effect problem where the change in perceived frequency, or pitch, is caused by the relative motion of the source of the sound in relation to the receiver (listener) of the sound.
Assuming for this problem that the convertibles with the trumpet players are moving apart and then together along the same line, then can calculate the perceived frequency (pitch) using the Doppler effect equation:
fp = fo (v + vr)/(v + vs), where
fp = perceived frequency
fo = original frequency
v = velocity of sound waves (in the medium, which is this case is air)
vs = velocity of the source
vr = velocity of the receiver relative to the medium
Given
fo = 1024 Hz
v = 343 m/s (speed of sound waves in air)
vs = +30 m/s or -30 m/s depending if the source is moving towards or away from the receiver
vr = 0 m/s/, +30 m/s or - 30 m/s depending if the receiver is standing still, moving towards or away from the source
For case A: the source and receiver are moving in opposite directions
vs = -30 m/s because it is moving away from the receiver
vr = +30 m/s in the same direction as the sound waves
fp = 1024 Hz (343m/s - 30 m/s)/(343 m/s + 30 m/s) = 859.3 Hz (lower pitch, or perceived frequency as both moving away)
For case B: the receiver is standing still and the source is moving away
vs = -30 m/s, vr = 0 m/s
fp = 1024 Hz (343 m/s - 30 m/s)/343 = 934.4Hz (lower pitch, or perceived frequency as just the source is moving away relative to the receiver)
For case C: both the receiver and source are moving towards each other
fp = 1024Hz (343 m/s + 30 m/s)/343 m/s - 30m/s) = 1220.3Hz (higher pitch, or perceived frequency)
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Stanton D.
Problem does not state that the convertibles were initially moving apart *along a common line connecting them*, only that the direction vectors of motion are in opposite directions. So the problem is unspecified, unless you add that stipulation, for parts (a) and (b). (Part (c) does specify "toward each other"). You can derive the formula relating perceived frequency, speed of sound in air, vehicle movement by considering how the wavefront moves vs. how the vehicle moves, to lay the wavefront crests along in space at a different spacing. Just don't forget that the other vehicle is moving, too, so the frequency shift at the other vehicle is twice what it is to the stationary observer. All of which should tell you, that there is a distinctive profile to the Doppler shift vs. time, which can tell you how close the straight route of passage is *to you*. I *think* you can get this independent of the vehicle speed and frequency of the emitted note. Maybe police radar works the same way, since the detector doesn't need to be reading directly head-on to your path. --Cheers, --Mr. d.03/30/22