Home Page Textbooks Physics Page 475: Lesson Check

Physics

1st Edition

Walker

ISBN: 9780133256925

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Page 475: Lesson Check

Exercise 39

Step 1

1 of 2

The motion is perpendicular to the direction of propagation, hence it is transverse wave

Result

2 of 2

See work here

Exercise 40

Step 1

1 of 2

We know that the speed of a wave traveling through a material is determined by the properties of that material.

In general we can say that waves travel faster in materials that are hard.

Since a chocolate bar is harder than a chocolate cake we conclude that the speed of a wave will be greater in the bar than it is in the cake.

Result

2 of 2

The speed of a sound wave will be greater in the chocolate bar.

Exercise 41

Step 1

1 of 2

Light waves carry electromagnetic energy. The energy coming from Light bulbs, lamps, LEDs etc., help us to see things. This energy falls in our retina and our brain detects that signal. Also using camera sensors or photographic films, we record or capture the information.

The energy coming from lasers is used for various purposes such as lasic surgery and laser cutting.

Radio waves also carry electromagnetic energy. This energy is used to send information one place to another such as FM programs, mobile phone calls, RADAR, etc. It is also used in navigation satellite.

Water waves carry mechanical energy. At the sea shores, we see lots of trash which is brought by the water waves (energy is used). There are some efforts being made to convert this water wave energy to the electricity as well.

Sound waves too carry mechanical energy in the form of pressure variations. This energy is recognized in form of pressure variation with the help of membrane such as eardrum. That is how we hear sound and also record sound using microphones.

Result

2 of 2

Light waves, radio waves, water waves and sound waves all carry the energy.

Exercise 42

Step 1

1 of 3

### Theoretical reminder

We know that period $T$ is defined as the time required to complete one full cycle of a given motion.
The frequency $f$ is the number of oscillations per unit time.

They are related by the following formula:

$$
begin{equation}
f = frac{1}{T}
end{equation}
$$

We know that the wavelength and frequency determine the speed of a wave. They are related by the following formula:

$$
begin{equation}
v = lambda cdot f
end{equation}
$$

Step 2

2 of 3

section*{Calculation}
begin{enumerate}[a)]
item
Our wave oscillates $n = 5$ times a second so the frequency is found directly:
begin{align*}
f = frac{n}{t} = frac{5}{1text{ s}} = 5text{ Hz}
end{align*}
item
The period is found using formula (1) as follows:
begin{align*}
T = frac{1}{f} = frac{1}{5text{ Hz}} = 0.2text{ s}
end{align*}
item
Since we know the frequency from part a, and we know the speed to be equal $v = 6.0 ; frac{text{m}}{text{s}}$, the wavelength of the wave can be determined using the formula (2) as follows:
begin{align*}
v = lambda cdot f \
end{align*}
Rearranging for $lambda$ we get:
begin{align*}
lambda = frac{v}{f}
end{align*}
Plugging in the numbers we get:
begin{align*}
lambda = frac{6.0 ; frac{text{m}}{text{s}}}{5text{ Hz}} = 1.2text{ m}
end{align*}
end{enumerate}

Result

3 of 3

begin{enumerate}[a)]
item
The frequency is $f = 5text{ Hz}$
item
The period is $T = 0.2text{ s}$
item
The wavelength of the wave is $lambda = 1.2text{ m}$
end{enumerate}

Exercise 43

Step 1

1 of 3

### Theoretical reminder

We know that period $T$ is defined as the time required to complete one full cycle of a given motion.
The frequency $f$ is the number of oscillations per unit time.

They are related by the following formula:

$$
begin{equation}
f = frac{1}{T}
end{equation}
$$

We know that the wavelength and frequency determine the speed of a wave. They are related by the following formula:

$$
begin{equation}
v = lambda cdot f
end{equation}
$$

Step 2

2 of 3

### Calculation

By counting the waves that pass by us we conclude that our wave oscillates $n = 12$ times during $t = 45text{ s}$ so the frequency is found directly:

$$
begin{align*}
f = frac{n}{t} = frac{12}{45text{ s}} approx 0.267text{ Hz}
end{align*}
$$

Now, since we know the frequency, and we know the wavelength (the distance from one crest to the next) to be equal $lambda = 7.5text{ m}$, the speed of the wave can be determined using the formula (2) as follows:

$$
begin{align*}
v = lambda cdot f = 7.5text{ m} cdot 0.267text{ Hz} approx 2 ; frac{text{m}}{text{s}}
end{align*}
$$

Result

3 of 3

The speed of the wave is $v = 2 ; frac{text{m}}{text{s}}$

Exercise 44

Step 1

1 of 3

We know that the wavelength and frequency determine the speed of a wave. They are related by the following formula:

$$
begin{equation}
v = lambda cdot f
end{equation}
$$

Step 2

2 of 3

### Calculation

We know that for the given wave, the distance from its crest to the next trough is $2.4text{ m}$.

This is half the distance from crest to crest, for which we know is equal to a wavelength, so we conclude:

$$
begin{align*}
lambda = 2 cdot 2.4text{ m} = 4.8text{ m}
end{align*}
$$

Since we now know the wavelength, and we know the speed to be equal $v = 5.6 ; frac{text{m}}{text{s}}$, the frequency of the wave can be determined using the formula (1) as follows:

$$
begin{align*}
v &= lambda cdot f \
f &= frac{v}{lambda}
end{align*}
$$

Plugging in the numbers we get:

$$
begin{align*}
f = frac{5.6 ; frac{text{m}}{text{s}}}{4.8text{ m}} approx 1.17text{ Hz}
end{align*}
$$

Result

3 of 3

The frequency of the wave is $f = 1.17text{ Hz}$

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