Question

A flutist assembles her flute in a room where the speed of sound is 342 m/sm/s . When she plays the note A, it is in perfect tune with a 440 Hz Hz tuning fork. After a few minutes, the air inside her flute has warmed to where the speed of sound is 346 m/sm/s .1.How many beats per second will she hear if she now plays the note A as the tuning fork is sounded?
2.How far does she need to extend the "tuning joint" of her flute to be in tune with the tuning fork?

Answer 1

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