Answer:
A) beats per second she will hear if she now plays the note A as the tuning fork is sounded = 5.13 beats per second
B) length she needs to extend the "tuning joint" of her flute to be in tune with the tuning fork = 0.0045m
Explanation:
A) First of all, wavelength = v/f
Where v is speed of wave and f is frequency.
Thus, wavelength of the sound wave of Note A is;
f2 = 440 Hz and v = 342m/s
λ = 342/440 = 0.7773m
Now, since the air inside the note was warmed after a while, the wave will will have a new frequency which we'll call (f1) and and new speed (v'), thus;
f2 = v'/λ = 346/0.7773 = 445.13 Hz
Now let's calculate beat frequency(fbeat).
fbeat = (f1 - f2)
So fbeat = 445.13 - 440 = 5.13Hz or 5.13 beats per second
B) Now, frequency of standing wave models (fm) = n(v/2L)
Where n is a positive integer and L is the open tube length
Making L the subject of the formula, we have; L = nv/2fm
Now from earlier derivation, we see that v = fλ and in this case, v=fλ
Thus, let's replace v with fλ to het;
L = nλ/2
If we take, n=1, L = (1 x 0.7773)/2 = 0.3887m
Now, when the air inside the tube has warmed, it will have a new length to eliminate beats and give same frequency of 440Hz.
So let's call this new length L1;
So L1 = v'/2(f2) = 346/(2x440) = 346/880 = 0.3932m
So the length she needs to extend the "tuning joint" of her flute to be in tune with the tuning fork will be;
ΔL = L1 - L = 0.3932 - 0.3887 = 0.0045m
