Name:
Student ID:
Brock University
Physics 1P22/1P92
Winter 2014
Dr. D’Agostino
Solutions for Tutorial 2: Travelling Waves and Sound (Chapter 15)
1. When water freezes, the density decreases and the bonds between molecules becomes
stronger. Do you expect the speed of sound to be greater in liquid water or in water
ice? Explain. [2 points]
Solution: It’s hard to say. The speed of a mechanical wave increases when the forces
between particles (i.e., the “stiffness” of the medium) increases, and the speed of a
mechanical wave decreases when the density of the medium increases. Ice is both
stiffer and denser than air, so it’s hard to answer the question without experimenting.
Looking at a table of sound speeds in various media indicates that sound speeds are
generally much greater in solids than in gases.
2. Consider a sinusoidal transverse wave described by
y = 17 cos (0.13x − 5.4t)
where x and y are measured in cm, and t is measured in seconds. Determine the
amplitude, wavelength, frequency, and speed of the wave. [4 points]
Solution: The amplitude of the wave can be read from the formula: A = 17 cm. The
wavelength can be determined as follows:
2π
= 0.13
λ
2π
λ=
0.13
λ = 48.33 cm
The frequency can be determined as follows:
2π
= 5.4
T
1
5.4
=
T
2π
5.4
f=
2π
f = 0.86 Hz
The speed of the wave is
v = fλ
5.4
2π
v=
2π
0.13
5.4
v=
0.13
v = 41.5 cm/s
3. The intensity of electromagnetic waves from the Sun is 1.4 kW/m2 just above the
Earth’s atmosphere. Determine the intensity of electromagnetic waves at the surface
of the Sun. (The radius of the Sun is about 6.96 × 108 m, the radius of the Earth is
about 6.37 × 106 m, and the distance between Earth and Sun is about 1.50 × 1011 m.)
[2 points]
Solution: By the time electromagnetic waves from the Sun reach the Earth, they have
spread over a spherical surface that has radius equal to the distance between the Sun
and the Earth. The electromagnetic waves from the Sun originate on the spherical
surface of the Sun. Thus, if P is the total power of the wave, rS is the radius of the
Sun, and R is the distance between Earth and Sun, then
IS =
P
4πrS2
and
IE =
P
4πR2
Therefore,
IS
=
IE
P
2
4πrS
P
4πR2
IS
P
4πR2
=
·
IE
4πrS2
P
2
IS
R
= 2
IE
r
S IS
R
=
IE
rS
2
R
IS =
IE
rS
2
1.50 × 1011 m
IS =
· 1.4 × 103 W/m2
8
6.96 × 10 m
IS = 6.5 × 107 W/m2
Note that the radius of the Earth is irrelevant information for this problem.
4. You hear a sound of a certain intensity and your ear senses a loudness level of 72 dB.
A second sound has an intensity that is 25 times as great as the first sound. Determine
the loudness level (in dB) of the second sound. [2 points]
Solution: Applying the formula
β = (10 dB) log10
I
I0
to each sound, we obtain
I2
I1
β2 − β1 = (10 dB) log10
− (10 dB) log10
I
I
0 0
I2
I1
β2 − β1 = (10 dB) log10
− log10
I0
I0
!
I
2
I0
I1
I0
β2 − β1 = (10 dB) log10
I2
β2 − β1 = (10 dB) log10
I1
β2 − β1 = (10 dB) log10 (25)
β2 − β1 = (10 dB) (1.3979)
β2 − 72 dB = 14 dB
β2 = 86 dB
5. A police radar unit uses electromagnetic waves of frequency 2.3 GHz. When these
radar waves are reflected from a car moving at 100 km/h towards the radar detector,
what frequency shift is measured? [2 points]
Solution: Using the formula
v0
v
and selecting the positive sign because the car is moving towards the radar detector,
we obtain
∆f = ±2f0
100 km/h
3 × 108 m/s
100, 000/3600 m/s
∆f = 2(2.3 × 109 Hz)
3 × 108 m/s
∆f = 426 Hz
∆f = 2(2.3 GHz)
The frequency shift is very small compared to the frequency of the radar waves. Detecting such small frequency shifts requires some technical tricks that you might like
to read about.