2020 A/L Physics Past Paper · Question 11 (English Medium)

Answer: (1) $2.5$ cm

Given information

We know:

Frequency $f = 300\ \text{Hz}$.

Speed $v = 30\ \text{m s}^{-1}$.

Two identical waves travel in opposite directions — superposing into a standing wave.

What the question asks

We want: the distance from a node to the antinode next to it, in cm.

Relevant theory

If the frequncy had been 16 Hz then all sixteen wavelengths would have been contained in v after 1s, and their wavelength would have been v/16.

Figure 3.8 represents two progressive waves of equal amplitude and frequency travelling in opposite directions.

Points such as N, called nodes of the stationary waves, where the displacement is always zero.

Points such as A with the greatest amplitude are called antinodes.

The wave length of a stationary wave is twice the distance between successive nodes or successive antinodes and equals the wavelength of either of the progressive waves.

Required equations

$v = f\lambda \Rightarrow \lambda = \frac{v}{f}$

$d_{NN} = \frac{\lambda}{2}$

$d_{NA} = \frac{d_{NN}}{2} = \frac{\lambda}{4}$

Applying it to this question

Plan:

First find the wavelength from $v = f\lambda$.

Then find the distance between successive nodes, $\lambda/2$.

The antinode sits halfway between them, so the node to the next antinode is $\lambda/4$.

Reasoning and calculation

$0.10$

$\lambda = \frac{30}{300}$

$10$

$\lambda = 0.10 \times 100$

$5.0$

$d_{NN} = \frac{10}{2}$

$2.5$

$d_{NA} = \frac{5.0}{2}$

Checks

Units: $\dfrac{\text{m s}^{-1}}{\text{s}^{-1}} = \text{m}$; $0.10\ \text{m} = 10\ \text{cm}$; everything after that stays in cm ✓.

Does it make sense? One wavelength (10 cm) holds two node-to-node spacings and four node-to-antinode spacings: $4 \times 2.5 = 10$ cm ✓.

$v / f / 4 * 100$

Conclusion

$d_{NA} = \frac{\lambda}{4} = 2.5\ \text{cm}$

$\lambda = 10$ cm.

Node to node $= 5$ cm.

Node to the next antinode $= \lambda/4 = 2.5$ cm.

The answer is (1).

Taking the node-to-antinode distance as $\lambda/2$.

$\lambda/2$ is node to the next node. The antinode sits halfway between them — $\lambda/4$.