| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2021 |
| Session | October |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Simple Harmonic Motion |
| Type | Prove SHM and find period: vertical spring/string (single attachment) |
| Difficulty | Standard +0.3 This is a standard SHM question requiring equilibrium analysis to find k, proving SHM by showing F ∝ -x, then applying standard SHM formulas for speed and time. All steps follow routine procedures taught in M3 with no novel insight required, making it slightly easier than average. |
| Spec | 4.10f Simple harmonic motion: x'' = -omega^2 x6.02g Hooke's law: T = k*x or T = lambda*x/l6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(mg = \frac{kmg}{l} \cdot \frac{2l}{5}\) | M1 | M1 for \(mg = T\) and use of Hooke's Law |
| \(k = \frac{5}{2}\) | A1* (2) | Given answer correctly obtained |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(mg - T = m\ddot{x}\) | M1 | Equation of motion, dim correct with all necessary terms; allow \(a\) for acceleration; condone sign errors |
| \(mg - \frac{5mg}{2l}\left(x + \frac{2l}{5}\right) = m\ddot{x}\) | DM1A1 | DM1 for equation with correct terms and use of Hooke's Law with variable extension measured from \(E\); needs \(\ddot{x}\) |
| \(-\frac{5g}{2l}x = \ddot{x}\), hence SHM | A1* (4) | Correct equation and conclusion |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(\omega = \sqrt{\frac{5g}{2l}}\); \(a = \frac{1}{4}l\) | B1ft; B1 | B1 ft for dimensionally correct \(\omega\) or \(\omega^2\); B1 for \(a = \frac{1}{4}l\) |
| \(v = a\omega = \frac{1}{4}l \times \sqrt{\frac{5g}{2l}}\) | M1 | Use of \(v = a\omega\) or \(v^2 = \omega^2(a^2 - x^2)\) with \(x=0\) |
| \(\frac{1}{4}\sqrt{\frac{5gl}{2}}\) oe | A1 (4) | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(\frac{1}{4} \times \frac{2\pi}{\omega}\) | M1 | M1 for use of \(\frac{1}{4} \times \frac{2\pi}{\omega}\) |
| \(\frac{\pi}{2}\sqrt{\frac{2l}{5g}}\) oe | A1ft (2) | cao |
# Question 3:
## Part (a):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $mg = \frac{kmg}{l} \cdot \frac{2l}{5}$ | M1 | M1 for $mg = T$ and use of Hooke's Law |
| $k = \frac{5}{2}$ | A1* (2) | Given answer correctly obtained |
## Part (b):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $mg - T = m\ddot{x}$ | M1 | Equation of motion, dim correct with all necessary terms; allow $a$ for acceleration; condone sign errors |
| $mg - \frac{5mg}{2l}\left(x + \frac{2l}{5}\right) = m\ddot{x}$ | DM1A1 | DM1 for equation with correct terms and use of Hooke's Law with variable extension measured from $E$; needs $\ddot{x}$ |
| $-\frac{5g}{2l}x = \ddot{x}$, hence SHM | A1* (4) | Correct equation and conclusion |
## Part (c):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $\omega = \sqrt{\frac{5g}{2l}}$; $a = \frac{1}{4}l$ | B1ft; B1 | B1 ft for dimensionally correct $\omega$ or $\omega^2$; B1 for $a = \frac{1}{4}l$ |
| $v = a\omega = \frac{1}{4}l \times \sqrt{\frac{5g}{2l}}$ | M1 | Use of $v = a\omega$ or $v^2 = \omega^2(a^2 - x^2)$ with $x=0$ |
| $\frac{1}{4}\sqrt{\frac{5gl}{2}}$ oe | A1 (4) | cao |
## Part (d):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $\frac{1}{4} \times \frac{2\pi}{\omega}$ | M1 | M1 for use of $\frac{1}{4} \times \frac{2\pi}{\omega}$ |
| $\frac{\pi}{2}\sqrt{\frac{2l}{5g}}$ oe | A1ft (2) | cao |
---3.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{9777abb8-a564-40d5-8d96-d5649913737b-08_307_437_244_756}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
A particle $P$ of mass $m$ is attached to one end of a light elastic spring of natural length $l$ and modulus of elasticity $k m g$, where $k$ is a constant. The other end of the spring is fixed to horizontal ground.
The particle $P$ rests in equilibrium, with the spring vertical, at the point $E$.\\
The point $E$ is at a height $\frac { 3 } { 5 } l$ above the ground, as shown in Figure 1.
\begin{enumerate}[label=(\alph*)]
\item Show that $k = \frac { 5 } { 2 }$
The particle $P$ is now moved a distance $\frac { 1 } { 4 } l$ vertically downwards from $E$ and released from rest. Air resistance is modelled as being negligible.
\item Show that $P$ moves with simple harmonic motion.
\item Find the speed of $P$ as it passes through $E$.
\item Find the time from the instant $P$ is released to the first instant it passes through $E$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2021 Q3 [12]}}