Wave interference worksheet with diagrams and questions on wave properties.

Let’s go through each question one by one, step by step.

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Question 1:

We are given a wave diagram and asked to find:
- The wavelength of the wave (in meters)
- The amplitude of the wave (in meters)

Looking at the wave:

→ Wavelength: This is the distance between two identical points on the wave — for example, from crest to crest or trough to trough. In the diagram, we can see that from point A to point C is one full wave cycle. The horizontal axis shows distance in meters. From A to C is labeled as “2 m”. So, wavelength = 2 meters.

→ Amplitude: This is the maximum height of the wave from the center line (equilibrium position) to a crest or trough. Looking at the vertical scale, the wave goes up to +0.5 m and down to -0.5 m. So amplitude = 0.5 meters.

✔ Answer for Q1: Wavelength = 2 m, Amplitude = 0.5 m

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Question 2:

This graph shows displacement vs time for a sound wave traveling at 340 m/s.

Part a: Find frequency.

Frequency = number of waves per second → look at how many complete cycles happen in 1 second.

From the graph: At t=0 to t=1 sec, there are 4 complete waves.

So frequency = 4 Hz

Part b: Find wavelength.

Use the formula:
Speed = Frequency × Wavelength
→ v = f × λ
→ λ = v / f
→ λ = 340 m/s ÷ 4 Hz = 85 meters

✔ Answer for Q2a: 4 Hz
✔ Answer for Q2b: 85 m

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Question 3:

Count nodes and antinodes in each standing wave diagram.

First diagram (left): Looks like one loop — it has 2 ends fixed (nodes), and 1 middle bump (antinode).

Wait — let’s be precise.

In standing waves:
- Nodes = points with no movement (usually at ends if fixed)
- Antinodes = points with max movement (middle of loops)

Left diagram: It’s shaped like a single oval — so it has:
- 2 nodes (at both ends)
- 1 antinode (in the center)

Middle diagram: Two loops side by side — so:
- 3 nodes (ends + middle where they meet)
- 2 antinodes (one in each loop)

Right diagram: Four loops — so:
- 5 nodes (ends + 3 in between)
- 4 antinodes (one per loop)

But wait — let me double-check standard patterns.

Actually, for standing waves on a string fixed at both ends:

Number of nodes = n + 1
Number of antinodes = n
Where n = harmonic number (number of loops)

Left: 1 loop → nodes = 2, antinodes = 1 ✔
Middle: 2 loops → nodes = 3, antinodes = 2 ✔
Right: 4 loops → nodes = 5, antinodes = 4 ✔

✔ Answer for Q3:
Left: Nodes = 2, Antinodes = 1
Middle: Nodes = 3, Antinodes = 2
Right: Nodes = 5, Antinodes = 4

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Question 4: True or False?

“Constructive interference occurs when a crest meets another crest at a given location along the medium.”

Yes — constructive interference happens when two crests (or two troughs) meet — they add up to make a bigger wave.

✔ TRUE

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Question 5: True or False?

“Destructive interference occurs when a trough meets another trough at a given location along the medium.”

No! Destructive interference happens when a crest meets a trough — they cancel out.

If two troughs meet, that’s actually constructive interference — they make a deeper trough.

✘ FALSE

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Question 6:

Two pulses shown approaching each other — one upward, one downward.

When they meet, since one is up and one is down, they will interfere destructively.

The resulting amplitude = difference of the two amplitudes.

Assuming both have same size (say, amplitude = 1 unit each), then result = 1 - 1 = 0 → flat line.

But looking at the diagram — left pulse is tall up, right pulse is shorter down? Wait — actually, in the image, both seem to be same height but opposite direction.

Standard case: if equal and opposite, they cancel completely.

Resulting shape: straight line (no displacement) during overlap.

After passing, they continue unchanged.

But the question asks: “what would be the resulting amplitude?”

At the moment they fully overlap, if they are equal and opposite, amplitude = 0.

✔ Answer: The resulting amplitude is zero (they cancel out).

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Now, compiling all answers clearly:

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Final Answer:

1. Wavelength = 2 m, Amplitude = 0.5 m
2. a) Frequency = 4 Hz; b) Wavelength = 85 m
3. Left: Nodes = 2, Antinodes = 1; Middle: Nodes = 3, Antinodes = 2; Right: Nodes = 5, Antinodes = 4
4. True
5. False
6. Resulting amplitude = 0 (complete cancellation)