2020 A/L Physics Past Paper · Question 09 (English Medium)

Answer: (1) $B$ and $F$

What the question asks

A displacement-against-distance graph at one instant (a snapshot) is given. The wave travels in the $+x$ direction. Among six particles, find the pair whose velocity (speed and direction) is equal.

Relevant theory

Crests and troughs thus formed travel horizontally, or in other words, along a direction perpendicular to the direction of motion of the surface of water.

The next particular particle begins to oscillate up and down slightly later than the one immediately before it.

Required equations

$y = a\sin(kx - \omega t)$

$v = \frac{\partial y}{\partial t} = -a\omega\cos(kx - \omega t)$

$v = -\frac{\omega}{k}\,\frac{\partial y}{\partial x}$

Reasoning and calculation

Where displacement is largest (crest, trough), the slope is zero.

So the velocity is zero there.

Where displacement is zero ($A$, $D$), the size of the velocity is largest ($a\omega$).

The zeros are at the crest ($\approx 0.25\lambda$) and the trough ($\approx 0.75\lambda$), where $v = 0$.

Around each zero, the direction of the velocity changes.

Every particle follows the one just before it.

As the wave moves right, each particle moves to the position the particle on its left has now.

Shift the wave slightly in the $+x$ direction to see the direction.

Moving down: $A$, $B$ (rising slope), $F$ (rising slope again).

Moving up: $C$, $D$, $E$ (falling slope).

$-0.82a\omega$

$v_B = -a\omega\cos(0.19\pi)$

$v_F = -a\omega\cos(1.81\pi)$

$1.62\pi$

$2\pi(0.906 - 0.096) = 1.62\pi$

$v_B = v_F$

Limit check: move $B$ toward $A$ ($\theta \to 0$); then $F$ moves toward the end ($1 - \theta \to 1$).

Both then reach $-a\omega$, just like $A$ — equal for any $\theta$ at this instant.

$B$ and $F$ are not antiphase: their displacements are opposite, but their velocities are equal.

Reading the options

The equal-velocity pairs are:

($B$, $F$) — both moving down at $0.82a\omega$.

($C$, $E$) — both moving up at $0.82a\omega$ (not offered as an option).

So the answer is (1).

Why the other options fail

$A$ down, $D$ up — equal speed, opposite velocity.

$B$ down, $C$ up — equal displacement, opposite velocity.

$C$ up, $F$ down.

$B$ down, $E$ up.

Conclusion

$B$ and $F$ are both on rising slopes of equal steepness; so both move down at $0.82a\omega$ — the answer is (1).

To find direction on a snapshot: redraw the wave shifted slightly in the direction of travel; particles whose new curve is higher are moving up.

Exam tip