It's a simple problem I think but I can't get the answers to match with the book.
A particle fingering flute sounds note with frequency of 880Hz at 20 deg celcius(343m/s => sound wave speed) The flute is open at both ends.
a) find the air column length
b) find the frequency the flute produces at begining of football game when ambient temp is -50.0C and muscician has not had time to warm up flute.
the answers are a) 0.195m and b) 841m but I can't seem to get the answers when I use the equation for the standing wave in a open pipe
Thank you
If it is open on both ends, it is a half wave length. lambda=343/880, so lambda/2 is ...
Ok, at 50C, the wave speed increases by .6*(50-20)....see the formula in your text...and air column length stays the same (lambda/2).
Thank you bobpursley I got it =D
To solve this problem, we can start by using the equation for the standing wave in an open pipe:
λ = 2L/n
where λ is the wavelength of the sound wave, L is the length of the air column, and n is the harmonic number. We know that the flute is open at both ends, so the first harmonic is the fundamental frequency, n = 1.
For part a) of the question, we need to find the length of the air column. We are given the frequency of the note (880 Hz) and the speed of sound in air at 20°C (343 m/s). We can start by finding the wavelength using the formula:
λ = v/f
where v is the speed of sound (343 m/s) and f is the frequency (880 Hz). Plugging in the values, we get:
λ = 343 m/s / 880 Hz ≈ 0.3902 m
Now, we can use the formula for an open pipe to find the length of the air column:
λ = 2L
Substituting in the value for λ, we have:
0.3902 m = 2L
Solving for L:
L = 0.3902 m / 2 ≈ 0.195 m
So, the length of the air column in the flute is approximately 0.195 m.
For part b) of the question, we need to find the frequency of the note when the ambient temperature is -50.0°C. Since the flute has not had time to warm up, the speed of sound will still be 343 m/s. We can repeat the same steps as above, but with the new temperature.
Using the same formula, we find the wavelength:
λ = 343 m/s / 880 Hz ≈ 0.3902 m
Now, we can use this wavelength to find the frequency at the new temperature using:
f = v/λ
Plugging in the values, we have:
f = 343 m/s / 0.3902 m ≈ 879.6 Hz
So, the frequency the flute produces at the beginning of the football game when the ambient temperature is -50.0°C and the musician has not had time to warm up the flute is approximately 879.6 Hz.
It seems that the answers you provided are slightly different from the calculated values. There could be rounding errors or discrepancies in the given answers.