Tenth grade Lesson Using the Wave Equation to Solve Problems - Free Printable

Let’s go through each problem one by one, step by step. I’ll make sure to check my math and logic carefully before giving the final answer.

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Problem 1: Draw a transverse wave and label amplitude, wavelength, crest, and trough.

This is a drawing task — since we can’t draw here, I’ll describe what you should draw:

- A transverse wave looks like a wavy line going up and down (like ocean waves or a rope shaken side to side).
- Label the crest: the highest point of the wave.
- Label the trough: the lowest point of the wave.
- Label the amplitude: the vertical distance from the middle (rest position) to the crest (or to the trough — they’re equal).
- Label the wavelength: the horizontal distance between two crests (or two troughs).

✔ You did this correctly in your image — good job!

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Problem 2: Draw a longitudinal wave and label compression, rarefaction, and wavelength.

Again, a drawing task — but here’s how to do it:

- A longitudinal wave looks like a slinky being pushed and pulled — coils are bunched together and spread apart.
- Label compression: where the coils are close together (high pressure).
- Label rarefaction: where the coils are spread out (low pressure).
- Label wavelength: the distance from one compression to the next compression (or one rarefaction to the next).

✔ Your drawing shows this well — nice work!

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Problem 3: A wave has wavelength of 15 m and speed of 300 m/s. What is the frequency?

We use the wave equation:

> Speed = Frequency × Wavelength
> or
> v = f × λ

We want to find frequency (f), so rearrange:

> f = v / λ

Plug in the numbers:

> f = 300 m/s ÷ 15 m
> f = 20 Hz

✔ Your calculation was correct — 20 Hz.

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Problem 4: Ms. Loky sees 10 waves pass in 30 seconds. Distance between peaks is 4 meters. Find frequency, period, wavelength, and speed.

Let’s break it down:

- Wavelength (λ): given as distance between successive peaks → 4 meters
- Frequency (f): number of waves per second → 10 waves / 30 seconds = 1/3 ≈ 0.333 Hz
- Period (T): time for one wave → total time / number of waves = 30 s / 10 = 3 seconds
- Also, T = 1/f → 1/(1/3) = 3 s ✔
- Speed (v): use v = f × λ → (1/3 Hz) × 4 m = 4/3 ≈ 1.333 m/s

Your answers:
- F = 0.33 Hz ✔
- P = 3 s ✔
- λ = 4 m ✔
- v = 1.33 m/s ✔

All correct!

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Problem 5: Which wave is high frequency? Low frequency? Relationship between wavelength and frequency?

Look at Wave A and Wave B.

- Wave A has more waves packed into the same space → shorter wavelength → higher frequency.
- Wave B has fewer waves → longer wavelength → lower frequency.

So:
- A = high frequency
- B = low frequency

Relationship:
When wavelength gets shorter, frequency gets higher (if speed stays the same).
They are inversely proportional.

You wrote: “wavelength & frequency are inversely prop.” — that’s exactly right!

✔ Correct reasoning and labeling.

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Problem 6: Three waves have same frequency, travel in different media. Rank speed from greatest to least. Explain.

Waves A, B, C — all same frequency.

But look at their wavelengths:

- Wave B has shortest wavelength (most squished)
- Wave C has medium wavelength
- Wave A has longest wavelength (stretched out)

Since speed = frequency × wavelength, and frequency is the same for all…

→ The wave with the longest wavelength has the highest speed.

So:

- Longest wavelength → fastest → Wave A
- Medium → Wave C
- Shortest → slowest → Wave B

Rank: A > C > B

You wrote: “A, C, B because shorter wavelength = slower speed” — which is correct logic!

✔ Perfect!

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Final Answer:
All problems solved correctly. For Problem 3: frequency = 20 Hz. For Problem 4: frequency = 0.33 Hz, period = 3 s, wavelength = 4 m, speed = 1.33 m/s. For Problem 5: A is high frequency, B is low; wavelength and frequency are inversely related. For Problem 6: speed ranking is A > C > B because longer wavelength means faster speed when frequency is constant.