Physics
Physics
1st Edition
Walker
ISBN: 9780133256925
Textbook solutions

All Solutions

Page 459: Practice Problems

Exercise 8
Step 1
1 of 3
### Theoretical reminder

For a mass $m$ attached to a spring with a spring constant $k$ we know that the period of simple harmonic oscillations will be:

$$
begin{equation}
T = 2 , pi ; sqrt{frac{m}{k}}
end{equation}
$$

Step 2
2 of 3
### Calculation

In our problem the mass attached to the spring increases by a factor of 4, so we can write for the old and new period:

$$
begin{align*}
T_1 &= 2 , pi ; sqrt{frac{m_1}{k}} \
T_2 &= 2 , pi ; sqrt{frac{m_2}{k}}
end{align*}
$$

Now inserting $m_2 = 4 , m_1$ we have:

$$
begin{align*}
T_2 = 2 , pi ; sqrt{frac{4 , m_1}{k}} = sqrt{4} cdot 2 , pi ; sqrt{frac{m_1}{k}} = 2 cdot T_1
end{align*}
$$

Result
3 of 3
The period increases by a factor of 2, that is $T_2 = 2 , T_1$
Exercise 9
Step 1
1 of 3
### Theoretical reminder

For a mass $m$ attached to a spring with a spring constant $k$ we know that the period of simple harmonic oscillations will be:

$$
begin{equation}
T = 2 , pi ; sqrt{frac{m}{k}}
end{equation}
$$

Step 2
2 of 3
### Calculation

In our problem both the mass attached to the spring and the spring constant increases by a factor of 2, so we can write:

For the old and new period:

$$
begin{align*}
T_1 &= 2 , pi ; sqrt{frac{m_1}{k_1}} \
T_2 &= 2 , pi ; sqrt{frac{m_2}{k_1}}
end{align*}
$$

Now inserting $m_2 = 2 , m_1$ and $k_2 = 2 , k_1$ we have:

$$
begin{align*}
T_2 = 2 , pi ; sqrt{frac{2 , m_1}{2k_1}} = 2pi ; sqrt{frac{m_1}{k}} = T_1
end{align*}
$$

Result
3 of 3
When both the mass and the spring constant are doubled the period remains the same, that is $T_2 = T_1$
Exercise 10
Step 1
1 of 4
### Theoretical reminder

For a mass $m$ attached to a spring with a spring constant $k$ we know that the period of simple harmonic oscillations will be:

$$
begin{equation}
T = 2 , pi ; sqrt{frac{m}{k}}
end{equation}
$$

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
### Knowns

The mass attached to our spring is $m = 0.32text{ kg}$

The frequency of our harmonic motion $f = 1.6text{ Hz}$

Step 3
3 of 4
### Calculation

First we will use the frequency-period relation (2) to find the period of our motion:

$$
begin{align*}
T = frac{1}{f} = frac{1}{1.6text{ Hz}} = 0.625text{ s}
end{align*}
$$

Now we will use formula (1) to find the spring constant $k$ using the mass of the spring, and the period of the motion.

Rearranging for the spring constant we have:

$$
begin{align*}
T &= 2 , pi ; sqrt{frac{m}{k}} \
T^2 &= left(2 , pi right)^2 ; frac{m}{k} \
k &= frac{4 , pi^2}{T^2} ; m
end{align*}
$$

Plugging in the values we get:

$$
begin{align*}
k = frac{4 , pi^2}{(0.625text{ s})^2 } cdot 0.32text{ kg} = 32.34 ; frac{text{N}}{text{m}}
end{align*}
$$

Result
4 of 4
The spring constant is $k = 32.34 ; frac{text{N}}{text{m}}$
Exercise 11
Step 1
1 of 4
### Theoretical reminder

For a mass $m$ attached to a spring with a spring constant $k$ we know that the period of simple harmonic oscillations will be:

$$
begin{equation}
T = 2 , pi ; sqrt{frac{m}{k}}
end{equation}
$$

Step 2
2 of 4
### Knowns

The spring constant of our spring is $k = 22 ; frac{text{N}}{text{m}}$

The period of our harmonic motion $T = 0.95text{ s}$

Step 3
3 of 4
### Calculation

We will use formula (1) to find the mass $m$ of our spring using the spring constant $k$ of the spring, and the period of the motion.

Rearranging for the mass we have:

$$
begin{align*}
T &= 2 , pi ; sqrt{frac{m}{k}} \
T^2 &= left(2 , pi right)^2 ; frac{m}{k} \
m &= frac{T^2}{(2 , pi)^2} ; k
end{align*}
$$

Plugging in the values we get:

$$
begin{align*}
m = frac{(0.95text{ s})^2 }{4 , pi^2 } cdot 22 ; frac{text{N}}{text{m}} approx 0.50text{ kg}
end{align*}
$$

Result
4 of 4
The mass of the spring is $m = 0.50text{ kg}$
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