1.repeat the previous exercise given that the car moves away from the wall.
2.a piano tuner uses a tuning fork to tune a string to 220 Hz.when the tension is 600 N,he hears a beat frequency of 2 Hz.the beat frequency increases if the tension is in-creased.what is the correct tension?
- 魏王將張遼Lv 71 十年前最愛解答
(1) Though I don't know what is the previous exercise, I guess that it is about the beat frequency for a car approaching/staying awaying from the wall as follows:
Applying the Doppler effect equation of apparent frequency:
where v = sound velocity in air
vs = velocity of the source (+ for staying away and - for approaching)
f = source frequency
So to speak, in the scenario, there is a source approaching the observer and a source staying away of the observer, both at velocity v.
Apparent frequency of the car itself to the observer f1= [u/(u-v)] × f
Apparent frequency of the car image to the observer f2= [u/(u+v)] × f
where u is the sound velocity in air.
Hence the beat frequency is given by:
(2) First of all, we deduce the expression for the piano frequency as follows:
Velocity of transverse wave on the string v = √(T/μ) where T and μ are tension and linear density of the string respectively.
Moreover, for a string vibrating at fundamental mode, wavelength λ = 2L
Hence the frequency is given by:
f = v/λ
So to speak, the frequency increases as the tension on the string increases.
And from the given, when tension is increased, the beat frequency increases, implying that the original frequency of the piano is already higher than the tuning fork, i.e. equals to 220 + 2 = 222 Hz
Hence, we have, when tension on the string is 600N, frequency = 222Hz
Therefore, in order to correct it back to 220 Hz, we use the following relation:
Putting △f = -2, f = 222 and T = 600, we have △T = -10.8 N
Therefore, the correct tension should be 600 - 10.8 = 589.2 N (Assuming that the string is still vibrating under the same mode as before).資料來源： My physics knowledge