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Attempts 127

A wave moves through its medium at 20 m/s with a wavelength of 4 m. What is the frequency of the wave?

  1. A
    5 s⁻¹
  2. B
    16 s⁻¹
  3. C
    24 s⁻¹
  4. D
    80 s⁻¹

Topic Flashcards

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Question

The 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.

Answer

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.

Question

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.

Answer

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.

Question

What is the period (T) of the wave in the original problem? How is the period related to the frequency you calculated (5 Hz)?

Answer

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.

Question

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?

Answer

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.

Question

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?

Answer

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.

Mini Quiz

1 / 3
Is the calculated frequency of 5 Hz a property of the wave itself, or does it depend on the observer's motion relative to the wave source?
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