Question

A buoy bobs up and down in the ocean. The waves have a wavelength of 2.5 meters, and they pass the buoy at a speed of 4.0 meters/second. What is the frequency of the waves?
1.6 Hertz.
10 Hertz.
7.5 Hertz.
4 Hertz.

Answer 1

Are you sure?

Answer 2

You are at a pool party and notice that 2 waves pass by a floating ball in 4 seconds. If the distance between each wave crest is 2 meters, what is the speed of the waves?
1 m/s.
6 m/s.
2 m/s.
1.5 m/s.

Answer 3

A student moves the end of a compressed coiled spring up and down to demonstrate wave motion. She then moves the compressed coiled spring faster but keeps everything else the same. Which characteristic will increase?

Answer 4

To find the frequency of the waves, we can use the formula:

\[
\text{Frequency} (f) = \frac{\text{Speed} (v)}{\text{Wavelength} (\lambda)}
\]

Given:
- Speed (\(v\)) = 4.0 meters/second
- Wavelength (\(\lambda\)) = 2.5 meters

Substituting the values into the formula:

\[
f = \frac{4.0 \, \text{m/s}}{2.5 \, \text{m}} = 1.6 \, \text{Hertz}
\]

Thus, the frequency of the waves is **1.6 Hertz**.

Answer 5

Let's double-check the calculation step by step.

We have:
- Wave speed (\(v\)) = 4.0 m/s
- Wavelength (\(\lambda\)) = 2.5 m

Using the formula for frequency:

\[
f = \frac{v}{\lambda}
\]

Let's substitute the values in:

\[
f = \frac{4.0 \, \text{m/s}}{2.5 \, \text{m}}
\]

Calculating that gives:

\[
f = \frac{4.0}{2.5} = 1.6 \, \text{Hertz}
\]

So, I can confirm that the frequency is indeed **1.6 Hertz**. Thank you for your patience!

Answer 6

To find the speed of the waves, we can use the formula:

\[
\text{Speed} (v) = \text{Frequency} (f) \times \text{Wavelength} (\lambda)
\]

First, we need to determine the frequency of the waves.

**Step 1: Calculate the frequency.**

You observe that 2 waves pass by in 4 seconds, so the frequency \(f\) can be calculated as follows:

\[
f = \frac{\text{Number of waves}}{\text{Time (seconds)}}
\]

\[
f = \frac{2 \, \text{waves}}{4 \, \text{seconds}} = 0.5 \, \text{Hertz}
\]

**Step 2: Use the wavelength.**

We are given that the distance between each wave crest (wavelength) is:

\[
\lambda = 2 \, \text{meters}
\]

**Step 3: Calculate the speed.**

Now we can calculate the speed of the waves:

\[
v = f \times \lambda
\]

\[
v = 0.5 \, \text{Hz} \times 2 \, \text{m} = 1 \, \text{m/s}
\]

Thus, the speed of the waves is **1 m/s**.

Answer 7

When the student moves the end of the compressed coiled spring faster while keeping everything else the same, the frequency of the wave will increase.

**Explanation:**
- **Frequency** is the number of waves that pass a point in a given amount of time. If the student moves the spring faster, more waves are created in the same period, resulting in a higher frequency.
- **Wavelength** will typically decrease as frequency increases (for a given wave speed), maintaining the wave speed constant.

Therefore, the characteristic that will increase is the **frequency** of the waves produced in the spring.