A wave moves through its medium at 20 m/s with a wavelength of 4 m. What is the frequency of the wave?
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A
5 s⁻¹
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B
16 s⁻¹
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C
24 s⁻¹
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D
80 s⁻¹
The frequency of the wave is 5 s⁻¹.
Given a wave speed of 20 m/s and a wavelength of 4 m, the frequency is calculated using the wave equation v=fλv=fλ, which yields 5 hertz, equivalent to 5 per second (s⁻¹).
A) 5 s⁻¹
This is correct. Rearranging the wave equation gives f=vλ=20 m/s4 m=5 s−1f=λv=4 m20 m/s=5 s−1. This means five complete wave cycles pass a point every second, which is consistent with the given speed and wavelength.
B) 16 s⁻¹
This appears to result from subtracting wavelength from speed (20 – 4 = 16), which has no physical meaning in wave mechanics. Frequency cannot be derived from arithmetic differences between speed and wavelength.
C) 24 s⁻¹
This likely comes from adding speed and wavelength (20 + 4 = 24), another invalid operation. Such a sum produces a number with incorrect units and no relevance to wave frequency.
D) 80 s⁻¹
This results from multiplying speed and wavelength (20 × 4 = 80), yielding units of m²/s, not s⁻¹. This confuses the wave equation with an undefined product and grossly overestimates the frequency.
Conclusion
Frequency is wave speed divided by wavelength, and only 5 s⁻¹ satisfies this relationship. The correct answer is A) 5 s⁻¹.

Topic Flashcards
Click to FlipThe wave equation is v = f × λ. Solve this equation for wavelength (λ) and then use it to find the wavelength of a sound wave with a frequency of 100 Hz traveling at 340 m/s.
λ = v / f = 340 m/s / 100 Hz = 3.4 m. This reinforces rearranging the fundamental wave equation and provides practice with typical sound wave values.
If the wave from the problem (v=20 m/s, λ=4 m) enters a new medium where its speed decreases to 10 m/s, what will be its new wavelength if the frequency remains unchanged? Explain why.
New λ = v_new / f. First, find the original frequency: f = v/λ = 20/4 = 5 Hz. The frequency is determined by the source and stays the same. Therefore, new λ = 10 m/s / 5 Hz = 2 m. This teaches the critical concept that wave speed and wavelength change when entering a new medium, but frequency is invariant.
What is the period (T) of the wave in the original problem? How is the period related to the frequency you calculated (5 Hz)?
Period T = 1 / f = 1 / 5 Hz = 0.2 seconds. The period is the time for one complete cycle to pass a point. This introduces the concept of period and its inverse relationship with frequency.
Using the wave speed (20 m/s) and frequency (5 Hz), how many complete wave cycles will pass by a fixed point in the medium over a duration of 3 seconds?
Number of cycles = frequency × time = 5 cycles/second × 3 seconds = 15 cycles. This connects frequency to a tangible, countable observation of wave cycles over time.
A different wave has a frequency of 0.25 Hz and a wavelength of 8 m. Calculate its speed and compare it to the speed from the original problem (20 m/s). What does this tell you about the relationship between these three quantities?
v = f × λ = 0.25 Hz × 8 m = 2 m/s. This wave is much slower. This shows that for a given wavelength, a lower frequency means a lower speed, and vice-versa. It emphasizes that wave speed is the product of its frequency and wavelength.