Physics question

fintans71

a whistle emitting a note of 4 kHz is whirled in a horizontal circle of radius 1 m at a constant speed. if the highest note heard by a person a large distance away is 4000 Hz find the speed of the whistle

Academia

Speed, v = distance / time

distance covered in circle = 2Pi*radius = (2)(22/7)(1) = 44/7 m

time = period = 1/frequency = (1/4000) s

Speed = (44/7) / (1/4000) m/s = (44/7)(4000/1) m/s = 25142.8571429 m/s

OR

Speed = 2Pi*r*f
Pi = 22/7, r = radius and f = frequency

Delphi91

Correct me if I'm wrong, but there seems to be some sort of error in the question.

The op states that the whistle emits a note of 4kHz and then goes on to state that "the highest frequency detected by a person a large distance away is 4000 Hz" - isn't that the same as 4 kHz? If the figures are true, then this would happen when the whistle was passing from left to right relative to the observer or similarly from right to left. It would also happen for all speeds of the whistle.

It's a practical example of the Doppler effect. The highest frequency occurs when the whistle, rotating in a horizontal circle is rotating towards the observer, the lowest frequency occurs when the whistle is rotating away from the observer.

Academia's solution gives you the speed that the whistle must be travelling at to complete a circle of radius 1m in 1/4000 seconds which is a totally different situation entirely. His figure of 25,142 m/s is 25.2 km/s or over twice the escape velocity from the earth - a tad unrealistic I think!

Try the following formula:
[Latex] \displaystyle f^{'}=f \left( \frac{v}{v-v_{s}} \right)[/Latex]

where [Latex]\displaystyle f^{'}[/Latex] = maximum frequency heard, [Latex]\displaystyle f[/Latex] = original frequency of the whistle, [Latex]\displaystyle v[/Latex] = speed of sound in air and [Latex]\displaystyle v_s[/Latex] = the velocity of the whistle when the highest frequency is heard - this is what you are trying to calculate.

Re-arrange the above formula to have [Latex]\displaystyle v_s[/Latex] as the subject, check your figures and try again....

Mr Pseudonym

Academia wrote: »

Speed, v = distance / time

distance covered in circle = 2Pi*radius = (2)(22/7)(1) = 44/7 m

time = period = 1/frequency = (1/4000) s

Speed = (44/7) / (1/4000) m/s = (44/7)(4000/1) m/s = 25142.8571429 m/s

OR

Speed = 2Pi*r*f
Pi = 22/7, r = radius and f = frequency

T = 1/4 000 gives the period of the sound wave, not of the circular motion of the whistle. Delphi91 gave the appropriate formula for this type of question.

Just to slightly clarify something Delphi91 said: the highest frequency occurs when the velocity of the whistle (tangent to orbit) is parallel to (and in the direction of) the observer.