Question

A flutist assembles her flute in a room where the speed of sound is 342m/s . When she plays the note A, it is in perfect tune with a 440Hz tuning fork. After a few minutes, the air inside her flute has warmed to where the speed of sound is 345m/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.3Hz

b.0.0034m

Explanation:

First, we know the flute is an open pipe, because open pipe as both end open and a close organ pipe as only one end close.

The formula relating the length and he frequency is giving as

$f=\frac{nv}{2l}$.

a.we first determine the length of the flute at the fundamental frequency i.e when n=1 and when the speed is in the 342m/s

Hence from

$f=\frac{nv}{2l}\\\\l=\frac{342}{2*440}\\ l=0.389m$.

since the value of the length will remain constant, we now use the value to determine the frequency when the air becomes hotter and the speed becomes 345m/s.

$f=\frac{nv}{2l} \\f=\frac{345}{2*0.389}\\f=443.4Hz$

Hence the require beat is

$B=/f_{1}-f_{2}/\\B=/440-443/\\B=3Hz$.

b. since the length is dependent also on the speed and frequency, we determine the new length when she plays with a fundamental frequency when the speed of sound is 345m/s

using the formula

$L_{new}=\frac{v}{2f}\\\\L_{new}=\frac{345}{2*440}\\\\L_{new}=0.39204$

Now to determine the extension,

$L_{extend}=L_{new}-L_{old}\\L_{extend}=0.39204- 0.38864\\L_{extend}=0.0034m$