Question 8
The acceleration (m s$^{-2}$) of a particle moving in simple harmonic motion is given by $a=-9x$, where $x$ is its displacement (m) from the origin.
The period of the motion in seconds is
Not quite. You took $\omega=9$ instead of $\omega^{2}=9$, giving $T=\frac{2\pi}{9}$.
That’s it. Step through the hints below to see why.
Not quite. You stopped at $\omega=3$. That is the angular frequency in rad s$^{-1}$, not the period in seconds.
Not quite. You multiplied instead of dividing, $T=2\pi\times\omega=2\pi\times 3=6\pi$.
The graph shows $x$ against $t$ with $\omega=3$. The orange part is the motion from $t=0$ up to the option’s value of $T$.
After $\frac{2\pi}{9}$ s the argument has gone up by only $\frac{2\pi}{3}$, a third of a cycle. The motion hasn’t repeated yet, so this is not the period.
After $\frac{2\pi}{3}$ s the argument has gone up by exactly $2\pi$. The motion repeats for the first time, so this is the period.
3 is $\omega$, in rad s$^{-1}$. After 3 s the motion is part way through its second cycle, so 3 s is not the period.
After $6\pi$ s the particle is back where it started, but it has done 9 full cycles. The period is the time for one.
Compare with the standard form of SHM.
$$a=\frac{d^{2}x}{dt^{2}}=-9x \quad\text{and}\quad \frac{d^{2}x}{dt^{2}}=-\omega^{2}x$$
Simple harmonic motion is defined by $\frac{d^{2}x}{dt^{2}}=-\omega^{2}x$, where $\omega>0$ is the angular frequency. The period formula is written in terms of $\omega$, so find $\omega$ first.
Match the coefficients, then solve for $\omega$.
$$\omega^{2}=9$$
$$\omega=\sqrt{9}=3$$
The constant multiplying $-x$ is $\omega^{2}$, not $\omega$.
$\omega$ is defined as positive, so take the positive square root.
Use the period formula.
$$T=\frac{2\pi}{\omega}=\frac{2\pi}{3}$$
The solution is $x=A\cos(\omega t+\alpha)$, and cosine repeats when its argument increases by $2\pi$. That happens when $t$ increases by $\frac{2\pi}{\omega}$.
Matching $-9x$ with $-\omega^{2}x$ gives $\omega^{2}=9$, so $\omega=3$. The period is $\dfrac{2\pi}{3}$ s.
$$T=\frac{2\pi}{\omega}=\frac{2\pi}{3}$$
Only mark this done when you could do it without help. Reading the solution does not count.
A Tangent Tuition mock exam question. It is not a past QCAA question, and Tangent Tuition is not affiliated with the QCAA.
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