Question

You make a simple instrument out of two tubes which looks like a flute and extends like a trombone. One tube is placed within another tube to extend the length of the instrument. There is an insignificantly small hole on the side of the instrument to blow on. Other than this, the instrument behaves as a tube with one end closed and one end open. You lengthen the instrument just enough so it emits a 1975Hz tone when you blow on it, which is the 4th allowed wave for this configuration. The speed of sound in air at STP is 343m/s.
The instrument's sound encounters warmer air which doubles its speed. What is the new amplitude of this sound, in m, if the original sound has an amplitude of 0.5m?

The instrument's sound encounters warmer air which doubles its speed. What is the new wavelength of this sound in m?

The instrument's sound encounters warmer air which doubles its speed. What is the new frequency of this sound in Hz?

What length do you have to extend the instrument to for it to make this sound in m?

What is the fundamental frequency of the instrument at this length in Hz?

Your friend has a similar instrument which plays a tone of 2034.25 Hz right next to yours so they interfere. What will the frequency of the combined tone be in Hz?

You play your instrument in a car travelling at 12m/s away from your friend travelling on a bicycle 4m/s away from you. What frequency does your friend hear in Hz?

Answer 1

Explanation:

We know

the frequency of the tube with one end open and the other end closed follows the given relations as

$f_{1}$ : $f_{2}$ : $f_{3}$ : $f_{4}$ = 1 : 3 : 5 : 7

∴ the 4th allowed wave is $f_{4}$ = 7 $f_{1}$

                                                                   = $\frac{7v}{4l}$

We know  $f_{4}$ = 1975 Hz and v = 343 m/s ( as given in question )

∴$l = \frac{7\times v}{4\times f_{4}}$

 $l = \frac{7\times 343}{4\times 1975}$

           = 0.303 m

We know that v = $f_{4}$ x $λ_{4}$

$\lambda _{4}= \frac{v}{f_{4}}$

$\lambda _{4}= \frac{343}{1975}$

                            = 0.17 m

Now when the warmer air is flowing, the speed gets doubled and the mean temperature increases. And as a result the wavelength increases but the amplitude and the frequency remains the same.

So we can write

v ∝ λ

or  $\frac{v_{1}}{v_{2}}= \frac{\lambda _{1}}{\lambda _{2}}$

Therefore, the wavelength becomes doubled = 0.17 x 2

                                                                             = 0.34 m

Now the new length of the air column becomes doubled

∴ $l^{'}$ = 0.3 x 2

                        = 0.6 m

∴ New speed, $v^{'}$ = 2 x 343

                                               = 686 m/s

∴ New frequency is $f^{'}=\frac{v^{'}}{4\times l^{'}}$

                                 $f^{'}=\frac{686}{4\times 0.6}$

                                               = 283 Hz

∴ The new frequency remains the same.

Now we know

$v_{s}$ = 12 m/s, $v_{o}$ = 4 m/s, $f_{o}$ = 1975 Hz

Therefore, apparent frequency is $f^{'}=f^{o}\left ( \frac{v+v_{s}}{v+v_{o}} \right )$

$f^{'}=1975\left ( \frac{343+12}{343+4} \right )$

            = 2020.5 Hz