Question 1
The acceleration $(\text{m}\,\text{s}^{-2})$ of an object moving with simple harmonic motion is modelled by $a=-2.95x$, where $x$ is its displacement $(\text{m})$ from the origin.
Determine the period of the motion in seconds.
Not quite. $1.72$ is $\sqrt{2.95}$, which is the angular frequency $n$. The period is $\frac{2\pi}{n}$. In 2023, 12.0% of Queensland students chose (A).
Not quite. $2.13$ is $\frac{2\pi}{2.95}$. In $a=-n^2x$ the number $2.95$ is $n^2$, so take the square root first. In 2023, 18.7% of Queensland students chose (B).
Not quite. $2.95$ is $n^2$, not the period. Find $n=\sqrt{2.95}$, then $T=\frac{2\pi}{n}$. In 2023, 12.2% of Queensland students chose (C).
That’s it. Step through the hints below to see why. In 2023, 56.2% of Queensland students chose (D).
A solution of $a=-2.95x$, starting at $x=1$. The dashed line is at the time in the option you test.
After this time the object is back at $x=1$ and about to repeat the motion. That is one full period.
The object isn’t back at its starting position yet, or it has gone past it, so this isn’t the period.
Simple harmonic motion has the form $a=-n^2x$. Match it with $a=-2.95x$.
$$n^2=2.95$$
Take the square root to get the angular frequency.
$$n=\sqrt{2.95}\approx1.718$$
The period is the time for one full cycle.
$$T=\frac{2\pi}{n}=\frac{2\pi}{\sqrt{2.95}}\approx3.66\ \text{s}$$
Compare $a=-2.95x$ with $a=-n^2x$, so $n=\sqrt{2.95}$.
$$T=\frac{2\pi}{n}=\frac{2\pi}{\sqrt{2.95}}\approx3.66\ \text{s}$$
How Queensland students answered in 2023
From the QCAA Specialist Mathematics subject report for 2023. The percentages can add to less than 100 because some students didn’t answer.
Only mark this done when you could do it without help. Reading the solution does not count.
Question wording is from the 2023 QCAA Specialist Mathematics external assessment, © State of Queensland (QCAA) 2023, licensed under CC BY 4.0, and has been adapted. Tangent Tuition is not affiliated with the QCAA.
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