| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2022 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Simple Harmonic Motion |
| Type | Prove motion is SHM from equation |
| Difficulty | Moderate -0.3 This is a straightforward SHM question requiring standard techniques: differentiate twice to show acceleration ∝ -displacement, read off period/amplitude from the given equation, and use inverse cosine for timing. All parts follow textbook procedures with no novel problem-solving required, making it slightly easier than average. |
| Spec | 4.10f Simple harmonic motion: x'' = -omega^2 x |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(\dot{x} = -4 \times \frac{\pi}{5}\sin\left(\frac{1}{5}\pi t\right)\) | M1 | Differentiate given expression for \(x\) twice; need cos→sin→cos |
| \(\ddot{x} = -4\times\left(\frac{\pi}{5}\right)^2\cos\left(\frac{1}{5}\pi t\right)\) | A1 | Both derivatives correct |
| \(\ddot{x} = -\left(\frac{\pi}{5}\right)^2 x\) ∴ SHM | A1 | Rewrite in standard SHM form and state conclusion |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| period \(= \frac{2\pi}{\frac{\pi}{5}} = 10\) (s) | M1A1 | Correct method; correct period |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| amplitude \(= 4\) (m) | B1 | Correct amplitude |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(\dot{x} = -4\times\frac{\pi}{5}\sin\left(\frac{1}{5}\pi t\right)\) or \( | \dot{x}_{\max} | = a\omega\) |
| Max speed \(= 4\times\frac{\pi}{5} = \frac{4\pi}{5}\) or \(0.8\pi\) (ms\(^{-1}\)) | A1 | Correct max speed |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| At \(A\): \(x=1.5\), \(1.5 = 4\cos\left(\frac{1}{5}\pi t\right) \Rightarrow t_A = \frac{5}{\pi}\cos^{-1}\left(\frac{1.5}{4}\right)\) | M1A1 | Find time from start to either \(A\) or \(B\); one correct time |
| At \(B\): \(x=-2.5\), \(-2.5 = 4\cos\left(\frac{1}{5}\pi t\right) \Rightarrow t_B = \frac{5}{\pi}\cos^{-1}\left(\frac{-2.5}{4}\right)\) | A1 | Second relevant time |
| Time \(A\) to \(B = t_B - t_A = \frac{5}{\pi}\cos^{-1}\left(\frac{-2.5}{4}\right) - \frac{5}{\pi}\cos^{-1}\left(\frac{1.5}{4}\right) = 1.6862... = 1.7\) or better (s) | A1 | Correct time from \(A\) to \(B\), 1.7(s) or better |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(1.5 = 4\sin\left(\frac{1}{5}\pi t\right) \Rightarrow t_A = \frac{5}{\pi}\sin^{-1}\left(\frac{1.5}{4}\right)\) | M1A1 | |
| \(2.5 = 4\sin\left(\frac{1}{5}\pi t\right) \Rightarrow t_B = \frac{5}{\pi}\sin^{-1}\left(\frac{2.5}{4}\right)\) | A1 | |
| \(_At_B = t_B + t_A = \frac{5}{\pi}\sin^{-1}\left(\frac{2.5}{4}\right) + \frac{5}{\pi}\sin^{-1}\left(\frac{1.5}{4}\right) = 1.6862...\) 1.7 or better | A1 |
# Question 5:
## Part (a):
| Working | Marks | Guidance |
|---------|-------|----------|
| $\dot{x} = -4 \times \frac{\pi}{5}\sin\left(\frac{1}{5}\pi t\right)$ | M1 | Differentiate given expression for $x$ twice; need cos→sin→cos |
| $\ddot{x} = -4\times\left(\frac{\pi}{5}\right)^2\cos\left(\frac{1}{5}\pi t\right)$ | A1 | Both derivatives correct |
| $\ddot{x} = -\left(\frac{\pi}{5}\right)^2 x$ ∴ SHM | A1 | Rewrite in standard SHM form and state conclusion |
## Part (b):
| Working | Marks | Guidance |
|---------|-------|----------|
| period $= \frac{2\pi}{\frac{\pi}{5}} = 10$ (s) | M1A1 | Correct method; correct period |
## Part (c):
| Working | Marks | Guidance |
|---------|-------|----------|
| amplitude $= 4$ (m) | B1 | Correct amplitude |
## Part (d):
| Working | Marks | Guidance |
|---------|-------|----------|
| $\dot{x} = -4\times\frac{\pi}{5}\sin\left(\frac{1}{5}\pi t\right)$ or $|\dot{x}_{\max}| = a\omega$ | M1 | Use either method to obtain max speed |
| Max speed $= 4\times\frac{\pi}{5} = \frac{4\pi}{5}$ or $0.8\pi$ (ms$^{-1}$) | A1 | Correct max speed |
## Part (e):
| Working | Marks | Guidance |
|---------|-------|----------|
| At $A$: $x=1.5$, $1.5 = 4\cos\left(\frac{1}{5}\pi t\right) \Rightarrow t_A = \frac{5}{\pi}\cos^{-1}\left(\frac{1.5}{4}\right)$ | M1A1 | Find time from start to either $A$ or $B$; one correct time |
| At $B$: $x=-2.5$, $-2.5 = 4\cos\left(\frac{1}{5}\pi t\right) \Rightarrow t_B = \frac{5}{\pi}\cos^{-1}\left(\frac{-2.5}{4}\right)$ | A1 | Second relevant time |
| Time $A$ to $B = t_B - t_A = \frac{5}{\pi}\cos^{-1}\left(\frac{-2.5}{4}\right) - \frac{5}{\pi}\cos^{-1}\left(\frac{1.5}{4}\right) = 1.6862... = 1.7$ or better (s) | A1 | Correct time from $A$ to $B$, 1.7(s) or better |
### ALT (e):
| Working | Marks | Guidance |
|---------|-------|----------|
| $1.5 = 4\sin\left(\frac{1}{5}\pi t\right) \Rightarrow t_A = \frac{5}{\pi}\sin^{-1}\left(\frac{1.5}{4}\right)$ | M1A1 | |
| $2.5 = 4\sin\left(\frac{1}{5}\pi t\right) \Rightarrow t_B = \frac{5}{\pi}\sin^{-1}\left(\frac{2.5}{4}\right)$ | A1 | |
| $_At_B = t_B + t_A = \frac{5}{\pi}\sin^{-1}\left(\frac{2.5}{4}\right) + \frac{5}{\pi}\sin^{-1}\left(\frac{1.5}{4}\right) = 1.6862...$ 1.7 or better | A1 | |
---\begin{enumerate}
\item A particle $P$ is moving along the $x$-axis. At time $t$ seconds the displacement of $P$ from the origin $O$ is $x$ metres, where $x = 4 \cos \left( \frac { 1 } { 5 } \pi t \right)$\\
(a) Prove that $P$ is moving with simple harmonic motion.\\
(b) Find the period of the motion.\\
(c) State the amplitude of the motion.\\
(d) Find, in terms of $\pi$, the maximum speed of $P$
\end{enumerate}
The points $A$ and $B$ lie on the $x$-axis, on opposite sides of $O$, with $O A = 1.5 \mathrm {~m}$ and $O B = 2.5 \mathrm {~m}$.\\
(e) Find the time taken by $P$ to move directly from $A$ to $B$.
\hfill \mbox{\textit{Edexcel M3 2022 Q5 [12]}}