A2 June 2024 Q5
5. A particle \(P\) moves in a straight line with simple harmonic motion about a fixed point \(O\).
The magnitude of the greatest acceleration of \(P\) is \(18\ \text{m s}^{-2}\)
When \(P\) is 0.3 m from \(O\), the speed of \(P\) is \(2.4\ \text{m s}^{-1}\)
The amplitude of the motion is \(a\) metres.
During one oscillation, the speed of \(P\) is at least \(2\ \text{m s}^{-1}\) for \(S\) seconds.
| Scheme | Marks | AO |
|---|---|---|
| \(a\omega^2 = 18\) | B1 | 3.4 |
| Use \(v^2 = \omega^2\left(a^2 - x^2\right)\) | M1 | 3.4 |
| \(2.4^2 = \omega^2\left(a^2 - 0.3^2\right)\) | A1 | 1.1b |
| Form equation in \(a\) only and solve for \(a\) e.g. \(\dfrac{a}{a^2 - 0.09} = \dfrac{18 \times 25}{144} \quad \left(18a^2 - 5.76a - 1.62 = 0\right)\) | M1 | 3.1a |
| \(\Rightarrow a = 0.5\) * | A1* | 1.1b |
| (5) |
Notes
B1: Use the model to state correct equation for greatest acceleration. Accept \(\pm 18\).
M1: Use the model to form a second equation in \(a\) and \(\omega\)
A1: Correct unsimplified equation with \(x\) and \(v\) substituted.
M1: Solve the simultaneous equations to obtain \(a\)
A1*: Correct only from correct working. Must have used positive \(\omega^2\).
| Scheme | Marks | AO |
|---|---|---|
| Solve for \(\omega\) and use max speed \(= a\omega\) | M1 | 3.4 |
| Greatest speed \(= 0.5 \times 6 = 3\ \left(\text{m s}^{-1}\right)\) | A1 | 1.1b |
| (2) |
Notes
M1: Complete method using the model to obtain greatest speed
E.g. \(v = \sqrt{\omega^2\left(0.5^2 - 0^2\right)}\)
A1: Correct only
| Scheme | Marks | AO |
|---|---|---|
| \(|v| = a\omega\sin\omega t\) OR \(v^2 = \omega^2\left(a^2 - x^2\right)\) and \(x = a\sin\omega t\) | B1ft | 3.3 |
| \(\Rightarrow 2 = 3\sin 6t \quad (t = 0.1216.....)\) OR \(\dfrac{\sqrt{5}}{6} = 0.5\sin 6t \quad (t = 0.14017\ldots)\) | M1 | 1.1b |
| \(S = \dfrac{2\pi}{6} - 4t\) OR \(S = 4 \times 0.14017\ldots\) | M1 | 3.1a |
| \(= 0.5607....\) | A1 | 1.1b |
| (4) | ||
| (11 marks) |
Notes
B1ft: Set up a correct model to find time when speed is 2. ft their \(\omega\)
(If starting from \(O\) then \(|v| = a\omega\cos\omega t\) )
M1: Solves their equation(s) to obtain a critical value for \(t\)
\((2 = 3\cos 6t \Rightarrow t = 0.14017...)\) (condone degrees: \(41.8^\circ/48.2^\circ\))
M1: Correct method to obtain \(S\); must be in radians \((S = 4 \times 0.14017...)\)
A1: 0.56 or better