Question

A flutist assembles her flute in a room where the speed of sound is 340 m/s. When she plays the note A, it is in perfect tune with a 440 Hz tuning fork. After a few minutes, the air inside her flute has warmed to where the speed of sound is 347 m/s .

Answer 1

Incomplete question as we have not told to find which quantity.The complete question is here

A flutist assembles her flute in a room where the speed of sound is 340 m/s. When she plays the note A, it is in perfect tune with a 440 Hz tuning fork. After a few minutes, the air inside her flute has warmed to where the speed of sound is 347 m/s .

(a)How many beats per second will she hear if she now plays the note A as the tuning fork is sounded?

(b) How far does she need to extend the "tuning joint" of her flute to be in tune with the tuning fork?

Answer:

(a)$f_{beat}=9.06Hz$

(b) The flutist extends to the turning joint of her flute by 1.363×10⁻⁵m

Explanation:

Part A

The wavelength of sound wave of the note A

λ=v/f

Where v speed of sound

f is frequency

Substitute the given values

So

λ=(340m/s)÷440Hz

λ=0.7727 m

The new frequency of note A when the air inside the flute has warmed

f¹=v¹/λ

f¹=(347m/s)/0.7727 m

f¹=449.05Hz

The Beat frequency of two waves is:

$f_{beat}=|f_{1}-f_{2} |\\f_{beat}=|449.05Hz-440Hz | \\f_{beat}=9.06Hz$

Part B

Frequencies of standing waves modes of an open tube of Length L:

$f_{m}=m(v/2L)\\where\\m=1,2,3,4........$

So

$L=m(v/2f_{m} )$

L=mλ/2

From part (a) The wavelength of sound of note A is 0.7727m and m=1

So

$L=(0.7727/2)=0.38635m$

When the air inside the flute has warmed.Then the new length is given below at same frequency of 440 Hz

So

$L_{a}=v/2f\\L_{a}=340m/s/2*440Hz\\L_{a}=0.38636m$

So the length the flutist extends to the turning joint for the flute to be:

ΔL=La-L

ΔL=0.38636m - 0.38635

ΔL=1.363×10⁻⁵m

The flutist extends to the turning joint of her flute by 1.363×10⁻⁵m

Answer 2

Answer:

Explanation:

Let L = length of flute used in playing A-note (open ends forming antinodes, L= lamda/2)

f1 = v1/2L = 440

340/2L = 440

2L = 340/440

= 0.386 m.

At v2 = 347 m/s,

f2 = v2/2L

= 347/2L

= 347 * 440/340

= 449.06 Hz

Beat frequency, f = f2 - f1

= 347 * 440/340 - 440

f = 9.06 beats/s

B.

f2 > f1 because now length is not able to keep ratio with increased speed of sound, so lets increase L to (L+x), f2' = f1 (again resonates) or beats f = 0

347/(2*(L + x)) = 440

2L + 2x = 347/440

340/440 + 2x = 347/440

2x = 7/440

x = 0.00795 m

= 7.95 mm.