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

a.3Hz

b.0.0034m

Step-by-step 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=(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=(nv)/(2l)\\\\l=(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=(nv)/(2l) \\f=(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)=(v)/(2f)\\\\L_(new)=(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\\`