Home Page Textbooks Physics Page 455: Practice Problems

Physics

1st Edition

Walker

ISBN: 9780133256925

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Page 455: Practice Problems

Exercise 1

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:

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

Step 2

2 of 3

The bird flaps its wings 5 times a second, so we can find the frequency directly from the definition as follows:

$$
begin{align*}
f = 5 ; frac{text{flaps}}{text{second}} = 5text{ Hz}
end{align*}
$$

The period we find by using formula (1):

$$
begin{align*}
T = frac{1}{f} = frac{1}{5text{ Hz}} = 0.2text{ s}
end{align*}
$$

Result

3 of 3

The frequency is $f = 5text{ Hz}$

The period is $T = 0.2text{ , }$

Exercise 2

Step 1

1 of 4

### 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:

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

Step 2

2 of 4

### General analysis

Using formula (1) we will analyze how the period changes when the frequency is modified.

When the frequency of the motion is doubled we can write $f_2 = 2 f_1$

It follows then that:

$$
begin{align*}
T_2 = frac{1}{f_2} = frac{1}{2 f_1} = frac{1}{2} cdot frac{1}{f_1} = frac{1}{2} cdot T_1
end{align*}
$$

That is, the period is halved.

Step 3

3 of 4

Now the bird flaps its wings 10 times a second, so we see that the frequency is doubled $f_2 = 10text{ Hz}$

The period we find by using formula (1):

$$
begin{align*}
T_2 = frac{1}{f_2} = frac{1}{10text{ Hz}} = 0.1text{ s}
end{align*}
$$

As expected the period is half of the one we would get for the bird flapping its wings 5 times a second. This result is in total agreement with the general analysis we did earlier.

Result

4 of 4

When the frequency is doubled the period is halved.

The period is $T = 0.1text{ s}$

Exercise 3

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:

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

Step 2

2 of 3

### Analysis

Using formula (1) we will analyze how the period changes when the frequency is modified.

Since $f cdot T = text{const}$, as $f$ increases it is obvious that $T$ decreases.

Hence we come to the conclusion that an increase in the frequency of oscillations will bring about a decrease in their period.

Result

3 of 3

When the frequency is increased the period decreases.

Exercise 4

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:

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

Step 2

2 of 3

### Calculation

A tennis ball is hit back and forth between two players. One full oscillation of this motion is characterized by the ball flying away from one player, being hit by the other, and returning to the first player. So the period is found by adding the time it takes flying in one direction and back:

$$
begin{align*}
T = 2 cdot 2.3text{ s} = 4.6text{ s}
end{align*}
$$

The frequency we find by using formula (1):

$$
begin{align*}
f = frac{1}{T} = frac{1}{4.6text{ s}} approx 0.22text{ Hz}
end{align*}
$$

Result

3 of 3

The period $T = 4.6text{ s}$

The frequency is $f = 0.22text{ Hz}$

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Page 455: Practice Problems

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