Moderate -0.5 This is a straightforward application of standard SHM formulas (v_max = ωa and a_max = ω²a) requiring only substitution of given values after calculating ω from the period. It's slightly easier than average as it involves direct formula recall with no problem-solving or interpretation needed.
1 A particle is moving with simple harmonic motion in a straight line. The period is 0.2 s and the amplitude of the motion is 0.3 m . Find the maximum speed and the maximum acceleration of the particle.
Maximum speed is \(0.3 \times 10\pi = 3\pi \approx 9.42 \text{ m s}^{-1}\)
M1
For relevant use of \(v = a\omega\)
A1\(\checkmark\)
For correct value \(3\pi\) or \(9.42\)
Maximum acceleration is \(0.3 \times (10\pi)^2 = 30\pi^2 \approx 296 \text{ m s}^{-2}\)
M1
For relevant use of \(a\omega^2\)
A1\(\checkmark\) [6]
For correct value \(30\pi\) or \(296\)
# Question 1:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $0.2 = \frac{2\pi}{\omega} \Rightarrow \omega = 10\pi$ | M1 | For relevant use of $\frac{2\pi}{\omega}$ |
| | A1 | For correct value $10\pi$ |
| Maximum speed is $0.3 \times 10\pi = 3\pi \approx 9.42 \text{ m s}^{-1}$ | M1 | For relevant use of $v = a\omega$ |
| | A1$\checkmark$ | For correct value $3\pi$ or $9.42$ |
| Maximum acceleration is $0.3 \times (10\pi)^2 = 30\pi^2 \approx 296 \text{ m s}^{-2}$ | M1 | For relevant use of $a\omega^2$ |
| | A1$\checkmark$ **[6]** | For correct value $30\pi$ or $296$ |
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1 A particle is moving with simple harmonic motion in a straight line. The period is 0.2 s and the amplitude of the motion is 0.3 m . Find the maximum speed and the maximum acceleration of the particle.
\hfill \mbox{\textit{OCR M3 Q1 [6]}}