Wave Speed Equation Practice Problems Answer Key
Introduction
Understanding wave speed is fundamental in the study of physics, particularly in the context of wave phenomena such as sound, light, and water waves. The wave speed equation provides a mathematical relationship that helps students and professionals analyze and predict wave behavior in various mediums. Practice problems are essential for mastering this concept, and having an answer key allows learners to check their work and deepen their understanding. This article offers a comprehensive set of practice problems related to the wave speed equation, complete with detailed solutions and explanations to serve as an answer key for learners.
The Wave Speed Equation
Before diving into practice problems, it is crucial to understand the core formula:
Wave speed (v) = Wavelength (λ) / Period (T)
Alternatively, the wave speed can be expressed in terms of frequency (f):
Wave speed (v) = Frequency (f) × Wavelength (λ)
Where:
- v is the wave speed (meters per second, m/s)
- λ (lambda) is the wavelength (meters, m)
- f is the frequency (hertz, Hz)
- T is the period (seconds, s)
Practicing with these equations helps reinforce understanding and prepares students for solving real-world problems involving wave motion.
Practice Problems with Solutions
Problem 1: Calculating Wave Speed from Wavelength and Frequency
A wave on a string has a wavelength of 0.5 meters and a frequency of 10 Hz. What is the wave speed?
Solution:
Using the formula:
v = f × λ
v = 10 Hz × 0.5 m
v = 5 m/s
Answer: The wave speed is 5 meters per second.
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Problem 2: Finding Wavelength Given Wave Speed and Frequency
A sound wave travels through the air at a speed of 340 m/s. If the frequency of the sound is 170 Hz, what is its wavelength?
Solution:
Rearranged formula:
λ = v / f
λ = 340 m/s / 170 Hz
λ = 2 m
Answer: The wavelength of the sound wave is 2 meters.
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Problem 3: Calculating Frequency from Wave Speed and Wavelength
A water wave has a wavelength of 2 meters and travels at a speed of 4 m/s. Find its frequency.
Solution:
Using the formula:
f = v / λ
f = 4 m/s / 2 m
f = 2 Hz
Answer: The frequency of the water wave is 2 Hz.
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Problem 4: Determining Wave Speed from Period
A wave has a period of 0.25 seconds and a wavelength of 1 meter. What is the wave speed?
Solution:
First, find the frequency:
f = 1 / T = 1 / 0.25 s = 4 Hz
Then, use the wave speed formula:
v = f × λ
v = 4 Hz × 1 m = 4 m/s
Answer: The wave speed is 4 meters per second.
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Problem 5: Complex Calculation — Multiple Variables
A wave on a string has a wave speed of 3 m/s, a wavelength of 0.6 meters. What is its frequency? Additionally, what is its period?
Solution:
Find the frequency:
f = v / λ = 3 m/s / 0.6 m = 5 Hz
Period T is the reciprocal of frequency:
T = 1 / f = 1 / 5 Hz = 0.2 seconds
Answer: The frequency is 5 Hz, and the period is 0.2 seconds.
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Practice Problems Summary
| Problem | Given | Find | Solution Formula | Final Answer |
|-----------|--------|-------|------------------|--------------|
| 1 | λ=0.5m, f=10Hz | v | v = f×λ | 5 m/s |
| 2 | v=340 m/s, f=170Hz | λ | λ = v / f | 2 m |
| 3 | λ=2m, v=4m/s | f | f = v / λ | 2 Hz |
| 4 | T=0.25s, λ=1m | v | v = f×λ, f=1/T | 4 m/s |
| 5 | v=3 m/s, λ=0.6m | f, T | f=v/λ, T=1/f | f=5Hz, T=0.2s |
Additional Practice Problems for Mastery
Problem 6: Real-World Application — Light Waves
The wavelength of visible light is approximately 500 nanometers (nm). If light travels at 3×10^8 m/s, what is its frequency?
Solution:
Convert wavelength to meters:
500 nm = 500 × 10^-9 m = 5×10^-7 m
Using the formula:
f = v / λ
f = 3×10^8 m/s / 5×10^-7 m = 6×10^14 Hz
Answer: The frequency is approximately 6×10^14 Hz.
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Problem 7: Sound Wave in Water
A sonar ping has a frequency of 25 kHz and travels through water at approximately 1500 m/s. What is its wavelength?
Solution:
Convert frequency to Hz:
25 kHz = 25,000 Hz
Calculate wavelength:
λ = v / f = 1500 m/s / 25,000 Hz = 0.06 m
Answer: The wavelength of the sonar ping is 0.06 meters.
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Problem 8: Unknown Variable — Solving for Wavelength
A wave with a speed of 600 m/s has a frequency of 300 Hz. Find its wavelength.
Solution:
λ = v / f = 600 m/s / 300 Hz = 2 m
Answer: The wavelength is 2 meters.
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Tips for Solving Wave Speed Problems
- Always identify which variables are given and which are to be found.
- Convert units where necessary to maintain consistency.
- Use the appropriate formula based on the given data.
- Remember that wave speed can be calculated directly if you have wavelength and frequency, or indirectly if you have wavelength and period.
Common Mistakes to Avoid
- Confusing period (T) and frequency (f). Remember T = 1 / f.
- Forgetting to convert units, especially with wavelengths in nanometers or micrometers.
- Misapplying formulas; ensure you use the correct relationship based on the problem.
Conclusion
Mastering wave speed calculations requires understanding the fundamental equations and practicing various problem types. The answer key provided here offers a range of problems from simple to complex, ensuring learners can verify their solutions and enhance their comprehension. Regular practice using these problems will build confidence and proficiency in analyzing wave phenomena across different contexts, from physics classrooms to real-world applications like telecommunications and oceanography.
Frequently Asked Questions
What is the formula for wave speed in terms of wavelength and frequency?
Wave speed (v) is calculated using the formula v = wavelength (λ) × frequency (f).
If a wave has a wavelength of 4 meters and a frequency of 3 Hz, what is its wave speed?
Using v = λ × f, the wave speed is v = 4 m × 3 Hz = 12 meters per second.
How do you solve for wavelength if you know the wave speed and frequency?
Rearranged from v = λ × f, the wavelength is λ = v / f.
A wave travels at 15 m/s and has a frequency of 5 Hz. What is its wavelength?
Wavelength λ = v / f = 15 m/s ÷ 5 Hz = 3 meters.
In a wave speed problem, if the wavelength is doubled and the frequency remains the same, how does the wave speed change?
Since wave speed v = λ × f, doubling the wavelength doubles the wave speed.
What is the significance of understanding the wave speed equation in real-world applications?
Understanding the wave speed equation helps in fields like communications, acoustics, and physics to analyze wave behavior and design effective systems.
Can wave speed be changed by altering the medium? Why or why not?
Yes, wave speed depends on the medium's properties; changing the medium's density or elasticity can increase or decrease the wave speed.
