Calculating Tune Adjustment for Flutist's Note A: Aa Flutist in 342 m/s Room

[sam.]

1. Homework Statement

Aa flutists assembles her flute in a room where the speed of sound is 342 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 346 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?

2. Homework Equations

For an open-open tube:

_m = 2L/m
f_m = mv/2L

3. The Attempt at a Solution

I found the answer for A to be 5 beats/second, but I can't seem to figure out how to calculate B. I tried subbing in f = 440 Hz and v = 346 m/s into the second equation with m=1 but it wasn't the right answer. I know the answer is 4.6 mm but I don't know how they get that. Any help is appreciated!

 

[Dick]

What's the difference in wavelength between a 440Hz wave at v=342m/sec and one at 346m/sec? How does this relate to having an open ended tube? Oh, and don't double post.

 

[ogb p]

I would approach this problem by first understanding the concept of beat frequency. When two sound waves with slightly different frequencies are played together, they will produce a beat frequency, which is the difference between the two frequencies. In this case, the flutist is playing a note with a frequency of 440 Hz, while the tuning fork has a frequency of 440 Hz. However, due to the change in the speed of sound, the flutist's note will now have a frequency of 442 Hz (346 m/s divided by 2L, where L is the length of the flute). This means that the beat frequency will be 2 Hz (442 Hz - 440 Hz).

To calculate the length of the flute needed to be in tune with the tuning fork, we can use the equation f_m = mv/2L, where f_m is the frequency of the note, m is the mode of the standing wave (in this case, m=1), v is the speed of sound, and L is the length of the flute. We know the frequency of the note (442 Hz) and the speed of sound (346 m/s), so we can rearrange the equation to solve for L.

L = v/(2f_m)

Substituting in the values, we get:

L = 346 m/s / (2*442 Hz) = 0.391 m = 39.1 cm

However, this is the length of the entire flute, including the tuning joint. To find the length of just the tuning joint, we need to subtract the length of the flute without the tuning joint (which is the length of the flute when the speed of sound is 342 m/s).

L_tuning joint = 39.1 cm - 34.2 cm = 4.9 cm

Therefore, the flutist needs to extend the tuning joint of her flute by 4.9 cm to be in tune with the tuning fork. This is approximately equal to 4.6 mm, as given in the problem.

In summary, as a scientist, I would approach this problem by understanding the concept of beat frequency and using the relevant equations to solve for the number of beats per second and the length of the tuning joint needed to be in tune with the tuning fork.