1. Music Examples Answers
1. A guitar string has a length of 0.80 m between the clamps.
(a) Draw the fundamental mode of vibration, and then find the
wavelength of the fundamental.
0.80 m
trough
crest
wavelength, λ
λ = 2 ( 0.80 m )
= m
A complete wave has a crest and a trough

2. 1. A guitar string has a length of 0.80 m between the clamps.
(b) If the waves travel at 512 m/s in the string, find the
fundamental frequency.
λ = 1.6 m v = 512 m/s
v = f λ
λ λ v
512 m/s
f = = λ
1.6 m
f = 320 Hz

3. 1. A guitar string has a length of 0.80 m between the clamps.
(c) Find the frequency of the sound wave produced by this string.
fundamental frequency (from part (b)) = 320 Hz
String vibrates 320 times per second
; so 320 times every second, string moves up, compressing the air above the
string
; so string creates 320 compressions per second
Sound frequency is the number of compressional waves
per second of the air
So the frequency of the sound wave produced
by this string = 320 Hz

4. 1. A guitar string has a length of 0.80 m between the clamps.
(d) If the air temperature is 24oC, find the wavelength of the
sound wave in the air.
f = 320 Hz λ = ?
v = f λ
v = ?
Need the speed of sound
in the air
For air, v = m/s + ( 0.6 ) T , where T = temperature in oC
v = m/s + ( 0.6 )( 24 )
= m/s
v = m/s

5. 1. A guitar string has a length of 0.80 m between the clamps.
(d) If the air temperature is 24oC, find the wavelength of the
sound wave in the air.
f = 320 Hz λ = ?
v = m/s
v = f λ
f f v
345.8 m/s
λ = =
Hz = 1/s f
320 Hz
λ = m

6. 1. A guitar string has a length of 0.80 m between the clamps.
(e) Find the wavelength and frequency of the second harmonic.
0.80 m
Fundamental
2nd Harmonic
There is one complete wave between the clamps, so λ = m
v = f λ
v = 512 m/s
λ λ v
512 m/s
f = =
= f = 640 Hz λ
0.80 m

7. 2. Find the wavelength (in the string) of the third longest traveling
wave that can set up a standing wave in a string of length 2.4 m.
Longest wave
2nd longest wave
3rd longest wave
3/2 wavelengths fit between clamps
2.4 m
2/3
3/2 λ = 2.4 m
( ) 2/3
λ = 1.6 m

8. 3. A closed tube 0.60 m long resonates on a 24oC day.
(a) Draw the tube and the longest wave that can resonate in this tube.
0.60 m
zoomed out
(b) Find the wavelength of the fundamental.
¼ of a wavelength is inside the tube (from the equilibrium
position to a crest)
4 ( ) 4
¼ λ = m
λ = 2.4 m

9. 3. A closed tube 0.60 m long resonates on a 24oC day.
(c) Find the fundamental frequency.
0.60 m
λ = 2.4 m
v = f λ
v = m/s ( on a 24oC day,
calculation done in Ex. #1d )
λ λ v
345.8 m/s
f = = λ
2.4 m
f = 144 Hz

10. 3. A closed tube 0.60 m long resonates on a 24oC day.
(d) Find the wavelength and frequency of the next longest wave
that can resonate in this tube.
0.60 m
zoomed out
¾ of a wavelength is inside the tube ( from the equilibrium
position to a crest through the equilibrium position
to a trough )
4/3 ( ) 4/3
¾ λ = m
λ = m

11. 3. A closed tube 0.60 m long resonates on a 24oC day.
(d) Find the wavelength and frequency of the next longest wave
that can resonate in this tube.
0.60 m
λ = m
f = ?
v = f λ
v = m/s ( on a 24oC day,
calculation done in Ex. #1d )
λ λ v
345.8 m/s
f = = λ
0.80 m
f = 432 Hz

12. 4. An open tube 0.60 m long resonates on a 24oC day.
(a) Draw the tube and the longest wave that can resonate in this tube.
0.60 m
zoomed out
(b) Find the wavelength of the fundamental.
½ of a wavelength is inside the tube (from a crest to a trough)
2 ( ) 2
½ λ = m
λ = 1.2 m

13. 4. An open tube 0.60 m long resonates on a 24oC day.
(c) Find the fundamental frequency.
0.60 m
λ = 1.2 m
v = f λ
v = m/s ( on a 24oC day,
calculation done in Ex. #1d )
λ λ v
345.8 m/s
f = = λ
1.2 m
f = 288 Hz

14. 4. An open tube 0.60 m long resonates on a 24oC day.
(d) Find the wavelength and frequency of the next longest wave
that can resonate in this tube.
0.60 m
zoomed out
Exactly one wavelength is inside the tube ( from one crest to
the next crest )
λ = m

15. 4. An open tube 0.60 m long resonates on a 24oC day.
(d) Find the wavelength and frequency of the next longest wave
that can resonate in this tube.
0.60 m
λ = m
f = ?
v = f λ
v = m/s ( on a 24oC day,
calculation done in Ex. #1d )
λ λ v
345.8 m/s
f = = λ
0.60 m
f = 576 Hz