| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2021 |
| Session | October |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Simple Harmonic Motion |
| Type | Find amplitude of SHM |
| Difficulty | Standard +0.3 This is a straightforward SHM problem requiring standard formulas (ω = 2π/T, v = ω√(a²-x²)) with given values. Part (a) involves algebraic manipulation to find amplitude, part (b) is direct recall that v_max = ωa. Slightly above average due to the algebraic work and SHM being a Further Maths topic, but follows standard textbook methods. |
| Spec | 4.10f Simple harmonic motion: x'' = -omega^2 x |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(\frac{2\pi}{\omega} = 4\) | M1 | Need to see this equation; allow with 4 or \(T\) in rearranged form |
| \(\omega = \frac{\pi}{2}\) | A1 | Seen |
| \(2 = a\frac{\pi}{2}\cos\left(\frac{\pi}{2} \times 0.5\right) \Rightarrow 2 = a\frac{\pi}{2} \times \frac{1}{\sqrt{2}}\) | M1 | Complete method to obtain equation in \(a\) only; use of \(x = a\sin\omega t\) to find \(x\) followed by \(v^2 = \omega^2(a^2 - x^2)\) may be seen; use of \(v = \pm a\omega\sin\omega t\) scores M0 |
| \(a = \frac{4\sqrt{2}}{\pi}\) m | A1* (4) | Correct answer correctly obtained |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(v_{MAX} = \frac{4\sqrt{2}}{\pi} \times \text{their } \omega\) | M1 | Use of \(a\omega\) with the given value of \(a\) |
| \(2\sqrt{2}\) (m s\(^{-1}\)) | A1 (2) | Allow 2.8 or better; ignore units but must be positive |
# Question 1:
## Part (a):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $\frac{2\pi}{\omega} = 4$ | M1 | Need to see this equation; allow with 4 or $T$ in rearranged form |
| $\omega = \frac{\pi}{2}$ | A1 | Seen |
| $2 = a\frac{\pi}{2}\cos\left(\frac{\pi}{2} \times 0.5\right) \Rightarrow 2 = a\frac{\pi}{2} \times \frac{1}{\sqrt{2}}$ | M1 | Complete method to obtain equation in $a$ only; use of $x = a\sin\omega t$ to find $x$ followed by $v^2 = \omega^2(a^2 - x^2)$ may be seen; use of $v = \pm a\omega\sin\omega t$ scores M0 |
| $a = \frac{4\sqrt{2}}{\pi}$ m | A1* (4) | Correct answer correctly obtained |
## Part (b):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $v_{MAX} = \frac{4\sqrt{2}}{\pi} \times \text{their } \omega$ | M1 | Use of $a\omega$ with the given value of $a$ |
| $2\sqrt{2}$ (m s$^{-1}$) | A1 (2) | Allow 2.8 or better; ignore units but must be positive |
---\begin{enumerate}
\item A particle $P$ is moving in a straight line with simple harmonic motion of period 4 s . The centre of the motion is the point $O$
\end{enumerate}
At time $t = 0 , P$ passes through $O$
At time $t = 0.5 \mathrm {~s} , P$ is moving with speed $2 \mathrm {~m} \mathrm {~s} ^ { - 1 }$\\
(a) Show that the amplitude of the motion is $\frac { 4 \sqrt { 2 } } { \pi } \mathrm {~m}$\\
(b) Find the maximum speed of $P$
\hfill \mbox{\textit{Edexcel M3 2021 Q1 [6]}}