Question

A musician in a concert hall is tuning her wind instrument. When she plays a short note she hears the echo of the note return from the opposite side of the 50.0 meter long auditorium 0.294 seconds later. Model the instrument as a tube closed at one end, if the instrument is properly tuned the note of the musician played would have a frequency of 233.082 Hz, but instead has a frequency of 226.513 Hz. This note is the first overtone, since the fundamental frequency is suppressed on this instrument. 1) What is the speed of sound in the concert hall on that day? 2) What is the temperature of the concert hall on that day in Kelvin? 3) What is the length of the tubing on the musician's instrument in meters? 4) How much does the musician need to change the length of the tubing to be in tune on that day? Does the tube need to be shortened or lengthened? Find the change in length of the tube, and make sure to use a negative number if the tube needs to be shorter. ( answer in meters.) 5) What is the fundamental frequency of this instrument when correctly tuned? 6) What is the second overtone of the correctly tuned instrument?  

Answer 1

Given data,

length of the hall. = 50 m

frequency of sound = 226.513 Hz.

The echo is heard after 0.294 s

1)

for the echo the sound has to travel twice the length

speed of sound = 2 * 50 / 0.294

= 340.14 m/s

speed of sound is proportional to T^1/2 , temperature

at STP T= 273K sound speed in air is 332 m/s

v / v_o = ( T / 273)^1/2

2)

temperature in hall on the specified day = 273 * (340.14 / 332)²  

= 286.5 K

3)

closed tube with fundamental frequency L = /4

with first overtone L = 3/4

v = f

= 4L/3 *f

length of the tube L = 3v / 4f

= 3 * 340.1 / 226.13 * 4

= 1.13 m

4)

correct frequency f = 233.082 Hz

= v / f

= 340.14 / 233.082

= 1.459 m

= 4L/3 -- first overtone

correct length of the tube L = 1.09 m

the tube has to be shortened by (1.09 - 1.13) = -0.04 m

5)

with fundamental frequency is

= 4L

= 1.09 * 4

= 4.36 m

f = v /

= 340.1 / 4.36

= 78.005 Hz

second overtone = 5f

= 5 ( 78.005)

= 392.023 Hz