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 .

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.

a

explanation:

because they are cartillidge

0.55x745.7j and a calculator