Question
A buoy bobs up and down in the ocean. The waves have a wavelength of 2.5 m, and they
pass the buoy at a speed of 4.0 m/s. What is the frequency of the waves? How much time
does it take for one wave to pass under the buoy?
pass the buoy at a speed of 4.0 m/s. What is the frequency of the waves? How much time
does it take for one wave to pass under the buoy?
Answers
Answer
no
Answered by
GPT-4o mini
To find the frequency of the waves, we can use the wave speed equation, which relates the speed (\(v\)), frequency (\(f\)), and wavelength (\(\lambda\)) of the wave:
\[
v = f \cdot \lambda
\]
Where:
- \(v\) is the speed of the wave (4.0 m/s)
- \(f\) is the frequency (which we are trying to find)
- \(\lambda\) is the wavelength (2.5 m)
Rearranging the equation to solve for frequency:
\[
f = \frac{v}{\lambda}
\]
Now, substituting in the values:
\[
f = \frac{4.0 \, \text{m/s}}{2.5 \, \text{m}} = 1.6 \, \text{Hz}
\]
So, the frequency of the waves is **1.6 Hz**.
Next, to find the time it takes for one wave to pass under the buoy, we can use the relationship between frequency and the period (\(T\)) of the wave:
\[
T = \frac{1}{f}
\]
Substituting the frequency we just calculated:
\[
T = \frac{1}{1.6 \, \text{Hz}} \approx 0.625 \, \text{s}
\]
Thus, it takes approximately **0.625 seconds** for one wave to pass under the buoy.
\[
v = f \cdot \lambda
\]
Where:
- \(v\) is the speed of the wave (4.0 m/s)
- \(f\) is the frequency (which we are trying to find)
- \(\lambda\) is the wavelength (2.5 m)
Rearranging the equation to solve for frequency:
\[
f = \frac{v}{\lambda}
\]
Now, substituting in the values:
\[
f = \frac{4.0 \, \text{m/s}}{2.5 \, \text{m}} = 1.6 \, \text{Hz}
\]
So, the frequency of the waves is **1.6 Hz**.
Next, to find the time it takes for one wave to pass under the buoy, we can use the relationship between frequency and the period (\(T\)) of the wave:
\[
T = \frac{1}{f}
\]
Substituting the frequency we just calculated:
\[
T = \frac{1}{1.6 \, \text{Hz}} \approx 0.625 \, \text{s}
\]
Thus, it takes approximately **0.625 seconds** for one wave to pass under the buoy.
Answered by
GPT-4o mini
I apologize if my response was not clear or if I misunderstood your query. Let’s go through the calculations again step by step.
### Given data:
- Wavelength (\(\lambda\)) = 2.5 m
- Wave speed (\(v\)) = 4.0 m/s
### Step 1: Calculate the Frequency
The formula to calculate the frequency (\(f\)) of a wave is:
\[
f = \frac{v}{\lambda}
\]
Substituting the given values:
\[
f = \frac{4.0 \, \text{m/s}}{2.5 \, \text{m}} = 1.6 \, \text{Hz}
\]
### Step 2: Calculate the Time Period
The time period (\(T\)) is the reciprocal of the frequency:
\[
T = \frac{1}{f}
\]
Substituting the frequency calculated:
\[
T = \frac{1}{1.6 \, \text{Hz}} \approx 0.625 \, \text{s}
\]
### Summary of Results:
- **Frequency of the waves**: 1.6 Hz
- **Time taken for one wave to pass**: approximately 0.625 seconds
If there are any additional points or a specific aspect you would like me to clarify or correct, please let me know!
### Given data:
- Wavelength (\(\lambda\)) = 2.5 m
- Wave speed (\(v\)) = 4.0 m/s
### Step 1: Calculate the Frequency
The formula to calculate the frequency (\(f\)) of a wave is:
\[
f = \frac{v}{\lambda}
\]
Substituting the given values:
\[
f = \frac{4.0 \, \text{m/s}}{2.5 \, \text{m}} = 1.6 \, \text{Hz}
\]
### Step 2: Calculate the Time Period
The time period (\(T\)) is the reciprocal of the frequency:
\[
T = \frac{1}{f}
\]
Substituting the frequency calculated:
\[
T = \frac{1}{1.6 \, \text{Hz}} \approx 0.625 \, \text{s}
\]
### Summary of Results:
- **Frequency of the waves**: 1.6 Hz
- **Time taken for one wave to pass**: approximately 0.625 seconds
If there are any additional points or a specific aspect you would like me to clarify or correct, please let me know!
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