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Physics

1st Edition

Walker

ISBN: 9780133256925

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Page 469: Lesson Check

Exercise 28

Step 1

1 of 3

### Theoretical reminder

The period of a pendulum oscillating due to gravity can be found using the length of the pendulum $L$ as follows:

$$
begin{equation}
T = 2 , pi sqrt{frac{L}{g}}
end{equation}
$$

Step 2

2 of 3

### Analysis

By looking at formula (1) we can see that the period $T$ of the pendulum is proportional to the square root of the length of the pendulum $L$, that is:

$$
begin{align*}
T sim sqrt{L}
end{align*}
$$

From this it is obvious that an increase in the length of the pendulum $L$ will lead to an increase in the pendulum’s period $T$.

Result

3 of 3

The period of the pendulum will increase if its length is increased.

Exercise 29

Step 1

1 of 2

From the definition of resonance, if the system (here, pendulum) is driven with its NATURAL FREQUENCY,it will start resonating.

Result

2 of 2

See work here

Exercise 30

Step 1

1 of 2

Pendulum A has larger amplitude than pendulum B.

We know that the maximal potential energy the pendulum has during its motion is when it is in the amplitude position.

So we conclude that pendulum A has larger maximal potential energy than pendulum B.

This means that pendulum A has larger total energy than pendulum B, since energy is conserved during the motion.

When expired as kinetic energy, we see that pendulum A has larger speed than pendulum B.

The same conclusion could have been reached simply by stating that pendulum A, in average, has to have greater speed than pendulum B because it has to move a larger distance in the same period of time.

Result

2 of 2

The average speed of pendulum A is greater than the average speed of pendulum B.

Exercise 31

Step 1

1 of 3

### Theoretical reminder

The period of a pendulum oscillating due to gravity can be found using the length of the pendulum $L$ as follows:

$$
begin{equation}
T = 2 , pi sqrt{frac{L}{g}}
end{equation}
$$

Step 2

2 of 3

### Calculation

We need to find the period of a swing of length $L = 3text{ m}$ at a location where the gravitational acceleration is $g = 9.81 ; frac{text{m}}{text{s}^2}$. We do this using formula (1):

$$
begin{align*}
T = 2 , pi sqrt{frac{L}{g}}
end{align*}
$$

Plugging in the numbers we get:

$$
begin{align*}
T &= 2 , pi sqrt{frac{3text{ m}}{9.81 ; frac{text{m}}{text{s}^2}}} \
T &= 3.47text{ s}
end{align*}
$$

Result

3 of 3

The period is $T=3.47text{ s}$

Exercise 32

Step 1

1 of 3

### Theoretical reminder

The period of a pendulum oscillating due to gravity can be found using the length of the pendulum $L$ as follows:

$$
begin{equation}
T = 2 , pi sqrt{frac{L}{g}}
end{equation}
$$

Step 2

2 of 3

### Calculation

We know that the period of the pendulum is $T = 1text{ s}$

We will use formula (1) to find the length of the pendulum.

First we need to rearrange it for $L$ as follows:

$$
begin{align*}
T &= 2 , pi sqrt{frac{L}{g}} \
T^2 &= 4 , pi^2 left(frac{L}{g} right) \
L &= left(frac{T}{2 , pi} right)^2 cdot g
end{align*}
$$

Plugging in the values we get:

$$
begin{align*}
L &= left(frac{1text{ s}}{2 , pi} right)^2 cdot 9.81 ; frac{text{m}}{text{s}^2} \
L &= 0.248text{ m} approx 0.25text{ m}
end{align*}
$$

Result

3 of 3

The length of the pendulum is approximately $L = 0.25 text{ m}$

Exercise 33

Step 1

1 of 3

### Theoretical reminder

The period of a pendulum oscillating due to gravity can be found using the length of the pendulum $L$ as follows:

$$
begin{equation}
T = 2 , pi sqrt{frac{L}{g}}
end{equation}
$$

Step 2

2 of 3

### Calculation

Firstly we find the period since we know that the pendulum performs 5 oscillations in 16s. This is done as follows:

$$
begin{align*}
T = frac{t}{n} = frac{16text{ s}}{5} = 3.2text{ s}
end{align*}
$$

Now to find the intensity of the gravitational acceleration at the location of the pendulum. This can be done using formula (1), since we know the length $L = 2.5text{ m}$ and the period $T = 3.2text{ s}$, as follows:

$$
begin{align*}
T = 2 , pi sqrt{frac{L}{g}}
end{align*}
$$

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

Plugging in the numbers we get:

$$
begin{align*}
g = frac{4 , pi^2 cdot 2.5text{ m}}{(3.2text{ s})^2} = 9.63 ; frac{text{m}}{text{s}^2}
end{align*}
$$

Result

3 of 3

$$
g = 9.63 ; frac{text{m}}{text{s}^2}
$$

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