C2. The Wave Model

Example 1

Problem: A transverse travelling wave has an amplitude $x_0$ and wavelength $\lambda$. What is the minimum distance between a crest and a trough measured in the direction of energy propagation?

• A. $2 x_0$
• B. $x_0$
• C. $\lambda$
• D. $\dfrac{\lambda}{2}$

Solution:

• The "direction of energy propagation" refers to the horizontal axis (distance).
• A full wave cycle spans from one crest to the next consecutive crest, which represents a total horizontal distance of $\lambda$.
• A trough occurs exactly halfway between two consecutive crests on the horizontal axis.
• Conclusion: The horizontal distance between a crest and a trough is exactly half a wavelength. The correct answer is D.

Example 2: Kinematics of Particles in a Wave

Problem: A wave on a string travels to the right as shown. The frequency of the wave is $f$. At time $t = 0$, a small marker on the string is in the position shown.

What is the position of the marker at $t = \dfrac{1}{4f}$?

Solution:

• Since Period $T = \dfrac{1}{f}$, the specified time $t = \dfrac{1}{4f}$ is exactly equal to $\dfrac{1}{4} T$ (one-quarter of a full wave period).
• In a transverse wave on a string, particles oscillate strictly up and down (perpendicular to the wave speed $v$). They do not travel horizontally.
• A particle starting at the maximum positive displacement (the crest) will take exactly one-quarter of a period to move downwards and reach the equilibrium position.

Example 3: Compressions in Longitudinal Waves

Problem: The graph shows the variation with distance $x$ of the displacement of the particles of a medium in which a longitudinal wave is travelling from left to right. Displacements to the right of equilibrium positions are positive. Which point is at the centre of a compression?

Solution:

• A compression is a region where particles crowd together, resulting in higher density.
• For particles to crowd into a specific location, particles to the left of that point must be displaced to the right (positive $y$ value), and particles to the right must be displaced to the left (negative $y$ value).
• Conclusion: The absolute centre of a compression is located where the graph crosses the x-axis (zero displacement) and the slope transitions from positive to negative values. This perfectly aligns with point B.

Example 4

Problem: The graphs show the variation of the displacement $y$ of a medium with distance $x$ and with time $t$ for a travelling wave. What is the speed of the wave?

Solution:

• Step 1: Extract $\lambda$. From the top $y-x$ graph, one full wave cycle spans from $0$ to $4.0\text{ cm}$. Therefore, $\lambda = 4.0\text{ cm}$.
• Step 2: Extract $T$. From the bottom $y-t$ graph, one full wave cycle spans from $0$ to $2.0\text{ ms}$. Therefore, $T = 2.0\text{ ms}$.
• Step 3: Apply the wave equation.
  $$v = \dfrac{\lambda}{T} = \dfrac{4.0\text{ cm}}{2.0\text{ ms}} = \mathbf{2.0 \text{ cm ms}^{-1}} = 20 \text{ m s}^{-1}$$

Example 5: Seismic Sound Waves Kinematics

Problem: Two sound waves from a point source on the ground travel through the ground to a detector. The speed of one wave is $7.5 \text{ km s}^{-1}$, the speed of the other wave is $5.0 \text{ km s}^{-1}$. The waves arrive at the detector $15$ seconds apart. What is the distance from the point source to the detector?

Solution:

• Step 1: Define time equations. Let distance be $D$. The time taken for each wave is $t = \dfrac{D}{v}$.
  Time for faster wave: $t_1 = \dfrac{D}{7.5}$
  Time for slower wave: $t_2 = \dfrac{D}{5.0}$
• Step 2: Form difference equation. The slower wave takes 15 seconds longer.
  $$\dfrac{D}{5.0} - \dfrac{D}{7.5} = 15$$
• Step 3: Solve for $D$. Multiply through by a common multiple (15):
  $$3D - 2D = 15 \times 15 \quad \Rightarrow \quad D = \mathbf{225 \text{ km}}$$