### 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:

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

### Analysis

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

Since:

$$
begin{align*}
f cdot T = text{const}
end{align*}
$$

As $T$ increases it is obvious that $f$ decreases.

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

When a particle is displaced from the equilibrium position, the force that wants to return it to that position is called the restoring force.

A crucial characteristic of the restoring force, needed for simple harmonic oscillations to occur, is the force’s linear proportionality to the displacement from equilibrium.

That is:

$$
begin{equation*}
F sim x
end{equation*}
$$

### 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:

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

### Calculation

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

$$
begin{align*}
T_1 &= 2 , pi ; sqrt{frac{m}{k_1}} \
T_2 &= 2 , pi ; sqrt{frac{m}{k_2}}
end{align*}
$$

Now inserting $k_2 = 4 , k_1$ we have:

$$
begin{align*}
T_2 = 2 , pi ; sqrt{frac{m}{4k_1}} = sqrt{frac{1}{4}} cdot 2 , pi ; sqrt{frac{m}{k_1}} = frac{1}{2} , T_1
end{align*}
$$

The period is halved, that is $T_2 = frac{T_1}{2}$

### 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:

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

### Analysis

To find the relation between the mass and the frequency we will combine formulas (1) and (2) as follows:

$$
begin{align*}
frac{1}{f} &= T = 2pi sqrt{frac{m}{k} } \
frac{1}{f} &= 2pi sqrt{frac{m}{k} }
end{align*}
$$

Now inverting for $f$:

$$
begin{align*}
f = frac{1}{2pi sqrt{frac{m}{k}}}
end{align*}
$$

We see that the frequency is inversely proportional to the square of the mass, that is:

$$
begin{align*}
f sim frac{1}{sqrt{m}}
end{align*}
$$

So we conclude that as the mass increases, the frequency will decrease.

### 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:

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

### Analysis

To find the relation between the mass and the frequency we will combine formulas (1) and (2) as follows:

$$
begin{align*}
frac{1}{f} &= T = 2pi sqrt{frac{m}{k} } \
frac{1}{f} &= 2pi sqrt{frac{m}{k} }
end{align*}
$$

Now inverting for $f$:

$$
begin{align*}
f = frac{1}{2pi sqrt{frac{m}{k}}}
end{align*}
$$

We see that the frequency is inversely proportional to the square of the mass, that is:

$$
begin{align*}
f sim frac{1}{sqrt{m}}
end{align*}
$$

So we conclude that to increase the frequency we should decrease the mass!

### 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:

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

### Knowns

– Masses: $m_{text{A}} = 0.1text{ kg}$ , $m_{text{B}} = 0.4text{ kg}$ , $m_{text{C}} = 0.4text{ kg}$ , $m_{text{D}} = 0.1text{ kg}$

– Spring constants: $k_{text{A}} = 10 ; frac{text{N}}{text{m}}$ , $k_{text{B}} = 40 ; frac{text{N}}{text{m}}$ , $k_{text{C}} = 10 ; frac{text{N}}{text{m}}$ , $k_{text{D}} = 40 ; frac{text{N}}{text{m}}$

We use formula (1) and the given values to calculate our period of all four mass-spring systems:

$$
begin{align*}
T_{text{A}} &= 2 , pi ; sqrt{frac{0.1text{ kg}}{10 ; frac{text{N}}{text{m}}}} = 0.63text{ s} \
T_{text{B}} &= 2 , pi ; sqrt{frac{0.4text{ kg}}{40 ; frac{text{N}}{text{m}}}} = 0.63text{ s} \
T_{text{C}} &= 2 , pi ; sqrt{frac{0.4text{ kg}}{10 ; frac{text{N}}{text{m}}}} = 1.3text{ s} \
T_{text{D}} &= 2 , pi ; sqrt{frac{0.1text{ kg}}{40 ; frac{text{N}}{text{m}}}} = 0.31text{ s} \
end{align*}
$$

Now we rank the systems in order of increasing period:

$$
begin{align*}
T_{text{D}} < T_{text{A}} = T_{text{B}} < T_{text{C}}
end{align*}
$$

The systems ranked in order of increasing period are:

$$
begin{align*}
D < A = B < C
end{align*}
$$

### 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:

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

When a spring is compressed or pulled by a given force $F$ its elongation $x$ is proportional to that force, and the coefficient is called the spring constant.

This is described by Hooke’s Law:

$$
begin{equation}
F = k , x
end{equation}
$$

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:

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

### Knowns

– The force stretching the spring $F = 12text{ N}$

– The amount the spring stretches $x = 0.16text{ m}$

– The mass attached to the spring $m = 2.2text{ kg}$

section*{Calculation}
begin{enumerate}[a)]
item
To find the period we will use relation (1). \
Before we do that we need to find the spring coefficient of stiffness $k$. We will do this using formula (2) as follows:
begin{equation*}
F = k , x
end{equation*}
Rearranging for $k$ we have:
begin{equation*}
k = frac{F}{x}
end{equation*}
Plugging in the values we get:
begin{align*}
k = frac{12text{ N}}{0.16text{ m}} = 75 ; frac{text{N}}{text{m}}
end{align*}
As mentioned earlier we now use formula (1) to find the period:
begin{equation*}
T = 2 , pi ; sqrt{frac{m}{k}}
end{equation*}
Plugging in the values we get:
begin{align*}
T = 2 , pi ; sqrt{frac{2.2text{ kg}}{75 ; frac{text{N}}{text{m}}}} = 1.077text{ s}
end{align*}
item
To find the frequency of oscillation we use formula (3) as follows:
begin{align*}
f = frac{1}{T} = frac{1}{1.07text{ s}} approx 0.93text{ Hz}
end{align*}
end{enumerate}

begin{enumerate}[a)]
item
The period is $T = 1.07text{ s}$
item
The frequency is $f = 0.93text{ Hz}$
end{enumerate}

### 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:

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

### Calculation

We know that the processor of a personal computer operates at $1.8 cdot 10^{9}text{ Hz}$.

The time required for one processing cycle can actually be found as the period for the given frequency of the computer processor.

Using formula (1) we have:

$$
begin{align*}
T = frac{1}{f} = frac{1}{1.8 cdot 10^{9}text{ Hz}} approx 0.56 cdot 10^{-9}text{ s}
end{align*}
$$

The time required for one processing cycle is $t = T = 0.56 cdot 10^{-9}text{ s}$