2020 A/L Physics Past Paper · Question 16 (English Medium)

Answer: (4) $\dfrac{a}{V}$

What the question asks

We want: the angular frequency $\omega$ from the largest velocity $V$ and the largest acceleration $a$ — an expression in $V$, $a$ (and $m$ if needed).

Relevant theory

v = ±w√( ( A² − x² ) ), where w = √(k/m)

a = −ω²x

Required equations

$v = \omega\sqrt{r^2 - x^2}$

$V = \omega r$

$a_x = -\omega^2 x$

$a = \omega^2 r$

Reasoning and calculation

Symbols: $r$ — amplitude; $\omega$ — angular frequency; $x$ — displacement from the centre; $V$ — largest speed (at the centre); $a$ — largest acceleration (at an end); $m$ — mass.

Eliminate the awkward amplitude $r$: dividing $a = \omega^2 r$ by $V = \omega r$ cancels $r$.

$\frac{a}{V} = \frac{\omega^2 r}{\omega r}$

$\omega = \frac{a}{V}$

Reading the options

Matching the form: $\omega = \dfrac{a}{V}$ — option (4).

In (2), $\dfrac{2\pi}{\omega} = T$.

In (3), an extra $2\pi$.

(5) is inverted.

(1) brings in an unneeded $m$.

T = 2π/ω

Checks

Dimension check: $\left[\dfrac{a}{V}\right] = \dfrac{LT^{-2}}{LT^{-1}} = T^{-1}$ — matches the angular frequency. $\left[\dfrac{V}{ma}\right] = M^{-1}T$ ✗, $\left[\dfrac{V}{a}\right] = T$ ✗.

Does it make sense? Doubling the amplitude $r$ doubles both $V$ and $a$; $\omega = a/V$ stays the same — matching the fact that SHM frequency does not depend on amplitude. With numbers: $r = 0.1$ m, $\omega = 4\ \text{rad s}^{-1}$ gives $V = 0.4$, $a = 1.6$, $a/V = 4$ ✓.

$a / V$

Conclusion

$\omega = \frac{a}{V}$

$V = \omega r$, $a = \omega^2 r$ ⇒ $\omega = \dfrac{a}{V}$ — the answer is (4).

Extra data like $m$ may be given just to test you; in mechanics questions, check with dimensions whether it is actually needed.

Exam tip