Problem 1. A person wears a hearing aid, which makes a sounds louder by 30 dB.
This means the sound intensity is increased times.
This means the amplitude of the sound-wave is increased times.
Solution. The number of decibels of difference between sound levels I
2and I
1is given by
2
1
10 log I β = ⎛ ⎜ I ⎞ ⎟
⎝ ⎠ . We can solve for the ratio I
2/I
1:
( )
2
1
30 3
2 10 10
1
log 10
10 10 10
I I I I
β
β
⎛ ⎞
⎜ ⎟ =
⎝ ⎠
= = =
The sound intensity increases by a factor of one thousand.
The intensity is proportional to the square of the amplitude. So, the amplitude increases only by a factor of 1000 = 31.6 . This means the amplitude of the sound-wave is increased 31.6 times.
I is proportional to A
2, In order to make I increase to 1000 times, A needs to increase to 1000 = 31.6 times.
A is proportional to √I
Attention: Please carefully review log functions. If log(x) = 2, x = 10
2;If log(y)=b , y= 10
bAlso, log(A/B) = log A- Log B Please check your old math book or found those online….
Problem 2. Two sound waves in air (call them X and Y) have wavelengths of 1.5 m and 3 m, and amplitudes of 0.01 Pa and 1 Pa, respectively.
a.) Find the frequency of each wave. f
X= Hz, f
Y= Hz Which sound wave has a higher pitch?
b.) Which sound wave is louder?
Fill in the following:
The intensity of Y is that of X by times.
Soundwave Y is dB soundwave X. (dB=decibels)
Solution. (a) We will use 343 m/s for the speed of sound. Then
( )
( )
( )
( )
343 m/s
229 Hz 1.5 m
343 m/s
114 Hz 3 m
X X
Y Y
f v
f v λ λ
= = =
= = =
Thus X has the higher pitch. Whichever has shorter wavelength has higher frequency.
(b) As for loudness, intensity and so forth, the wave with the higher pressure amplitude (Y) is louder.
Since intensity is proportional to the square of the amplitude, the intensity of Y is greater than that of
X by a factor of 100
2= 10,000. since each factor of 10 corresponds to 10dB, 10,000 = 10
4corresponds to 40 dB of difference.
Problem 3. Suppose you are stopped in your car at a traffic light, and an ambulance approaches you from behind with a speed of 14 m/s. The siren on an ambulance produces sound with a frequency of 800 Hz. What frequency will you hear? Hz
Solution. The observed frequency f' is given in terms of the source frequency f by
( ) ( )
( )
1 1
800 Hz 834 Hz
14 m/s
1 1
343 m/s
f f
u v
′ = = =
− −
Notice that the observer is not moving and the source is moving. You will hear 834Hz, which is higher than what is emitted. Conceptually this is because the wavelength is squeezed between the ambulance and you.
Decreased wavelength (negative sign of u/v at the bottom), gives you higher f.
Problem 4. What are the three lowest frequencies for standing waves on a wire 14.0 m long (fixed at both ends) having a mass of 88 g, which is stretched under a tension of 250 N?
Solution. We need the speed of waves in the wire,
( )( )
( )
:
250 N 14 m
199.4 m/s 0.088 kg
T TL
v = µ = m
= =
Then the three lowest frequencies are
( )
( )
1
2 1
3 1
199.4 m/s
7.12 Hz
2 2 14 m
f 2 14.24 Hz
f 3 21.36 Hz
f v L f f
= = =
= =
= =
In strings the speed of sound wave, v , is not 340m/s, but determined by the Tension and mass per unit length.
Strings with two fixed ends, have resonant wavelength of λ
Ν= 2L /N , and f
N= N v/2L, N are integers.
L
1 2 1
2
L f v
λ = = L
2 2
2 2
2 2
L v
f L
λ = =
3 2
2 3
3 2
L v
f L
λ = =
Problem 5. The human ear canal is much like an organ pipe that is closed at one end (at the tympanic
membrane, or eardrum) and open at the other (see the figure below). Suppose that a man's eardrum is of length L = 2.73 cm.
(a) What is the fundamental frequency and wavelength of the
ear canal? kHz cm
(b) Find the frequency and wavelength of the ear canal's third harmonic. (Recall that the third harmonic in this case is the standing wave with the second-lowest frequency.)
kHz cm
(c) Suppose a person has an ear canal that is longer than 2.73 cm. Is the fundamental frequency of that person's ear canal greater than, less than, or the same as the value found in part (a)?
Solution. (a) For pipes with one closed end, In the fundamental mode of resonance the wavelength is four times the length of the outer ear canal, and the corresponding frequency is calculated using the speed of sound in air.
( )
( )
1
1
4 4 2.73 cm 10.92 cm 343 m/s
3141 Hz 0.1092 m
L f v λ
λ
= = =
= = =
(b) The next resonance pattern has 3/4 of a wavelength inside the ear canal, and so has a
wavelength smaller by a factor of 1/3, and a frequency higher by a factor of 3:
( )
1
1
10.92 cm
3.64 cm 3
3141 Hz 3 9423 Hz
f x
λ = =
= =
(c) If the canal is longer, the resonance wavelengths increase, and the frequencies decrease.
Insight: at those high frequencies your ear canal can resonant with incoming sound with those frequency.
Would be uncomfortable. Luckily that’s not the frequency range of daily life sound or music.
(middle C is about 260 Hz)
Problem 6. A police car emitting a siren wail (450 Hz) is driving towards a fleeing suspect at v/35, where v is the speed of sound. The suspect is running away at v/45.
a.) What frequency does the suspect hear?
Hz
b.) To double-check your answer: is the distance between the car and suspect increasing or decreasing? Based on this, do you expect the frequency observed to be larger or smaller than that emitted? Is this what you got in part a?
Solution. : Now both the sound source and the observer are moving. The observer is moving away from the sound, the relative speed between observe and the sound wave is decreased. Like the kid running away from waves by the beech and he encounters less
4 4
L f v
λ= = L
L
f1
u1
f2 f3
u2
