| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2020 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Simple Harmonic Motion |
| Type | Complete motion cycle with slack phase |
| Difficulty | Challenging +1.2 This is a standard M3 SHM question requiring equilibrium analysis, verification of SHM conditions, and application of standard SHM formulas. Part (b) requires recognizing that the string never goes slack (since release point is below equilibrium), making it straightforward. Parts (c) and (d) are routine applications of v² = ω²(a² - x²) and time equations. The multi-part structure and need to work with modulus of elasticity elevates it slightly above average difficulty, but all techniques are standard textbook methods. |
| Spec | 4.10f Simple harmonic motion: x'' = -omega^2 x6.02g Hooke's law: T = k*x or T = lambda*x/l |
| VIXV SIHIANI III IM IONOO | VIAV SIHI NI JYHAM ION OO | VI4V SIHI NI JLIYM ION OO |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\dfrac{\frac{3}{4}mge}{a} = mg\) | M1 | Resolve vertically. Can be in either \(e\) or \(AP\). |
| \(e = \dfrac{4}{3}a \Rightarrow AP = \dfrac{7}{3}a\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(m\ddot{x} = mg - \dfrac{\frac{3}{4}mg\left(x+\frac{4}{3}a\right)}{a}\) | M1, A1 | Form equation of motion including weight and tension. Variable measured from equilibrium. A1: correct unsimplified equation. |
| \(m\ddot{x} = mg - \dfrac{3mgx}{4a} - mg = -\dfrac{3mgx}{4a}\) | DM1 | Solve to obtain \(\ddot{x} =\) . Must be \(\ddot{x}\) now. |
| \(\ddot{x} = -\dfrac{3g}{4a}x = -\omega^2 x \therefore\) SHM | A1 | Correct equation and statement. |
| \(\text{amp} = \dfrac{2}{3}a\), \(\text{KE} = -\dfrac{mg}{2}\) | B1 | Correct/sufficient values found to establish string remains taut. |
| \(\dfrac{2}{3}a < \dfrac{4}{3}a\) so string remains taut, therefore always SHM | B1 | Appropriate argument. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(v^2 = \dfrac{3g}{4a}\left(\left(\dfrac{2a}{3}\right)^2 - \left(\dfrac{a}{3}\right)^2\right)\) | M1, A1ft | Use of \(v^2 = \omega^2(a^2 - x^2)\) with their \(\omega\). ft on \(\omega\) and amplitude. |
| \(v^2 = \dfrac{ag}{4}\), \(v = \dfrac{\sqrt{ag}}{2}\) | A1 | Correct \(v\), c.s.o. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\dfrac{\frac{3mg}{4}(2a)^2}{2a} = \dfrac{\frac{3mg}{4}(a)^2}{2a} + mga + \dfrac{1}{2}mv^2\) | M1, A1 | Energy equation. Must contain 2 EPE terms, GPE and KE. |
| \(v = \dfrac{\sqrt{ag}}{2}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(-\dfrac{a}{3} = \dfrac{2a}{3}\cos(\omega t)\) or \(\dfrac{a}{3} = \dfrac{2a}{3}\sin(\omega t)\), \(\dfrac{T}{4} = \pi\sqrt{\dfrac{a}{3g}}\) | M1, A1ft | Use \(x = "a"\cos(\omega t)\) or \(x="a"\sin(\omega t)\). Amplitude consistent with their \(AP\). A1: correct equation in \(\omega\). |
| \(\cos(\omega t) = -\dfrac{1}{2}\) | ||
| \(t\sqrt{\dfrac{3g}{4a}} = \dfrac{2\pi}{3}\) | DM1, A1 | Solve to find \(t\). Must have full method for complete time. |
| \(t = \dfrac{2\pi}{3}\sqrt{\dfrac{4a}{3g}}\) | Any correct equivalent. Must come from correct working throughout; condone sign error in (b). |
# Question 6(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\dfrac{\frac{3}{4}mge}{a} = mg$ | M1 | Resolve vertically. Can be in either $e$ or $AP$. |
| $e = \dfrac{4}{3}a \Rightarrow AP = \dfrac{7}{3}a$ | A1 | |
---
# Question 6(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $m\ddot{x} = mg - \dfrac{\frac{3}{4}mg\left(x+\frac{4}{3}a\right)}{a}$ | M1, A1 | Form equation of motion including weight and tension. Variable measured from equilibrium. A1: correct unsimplified equation. |
| $m\ddot{x} = mg - \dfrac{3mgx}{4a} - mg = -\dfrac{3mgx}{4a}$ | DM1 | Solve to obtain $\ddot{x} =$ . Must be $\ddot{x}$ now. |
| $\ddot{x} = -\dfrac{3g}{4a}x = -\omega^2 x \therefore$ SHM | A1 | Correct equation **and** statement. |
| $\text{amp} = \dfrac{2}{3}a$, $\text{KE} = -\dfrac{mg}{2}$ | B1 | Correct/sufficient values found to establish string remains taut. |
| $\dfrac{2}{3}a < \dfrac{4}{3}a$ so string remains taut, therefore always SHM | B1 | Appropriate argument. |
---
# Question 6(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $v^2 = \dfrac{3g}{4a}\left(\left(\dfrac{2a}{3}\right)^2 - \left(\dfrac{a}{3}\right)^2\right)$ | M1, A1ft | Use of $v^2 = \omega^2(a^2 - x^2)$ with their $\omega$. ft on $\omega$ and amplitude. |
| $v^2 = \dfrac{ag}{4}$, $v = \dfrac{\sqrt{ag}}{2}$ | A1 | Correct $v$, c.s.o. |
**Alt:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\dfrac{\frac{3mg}{4}(2a)^2}{2a} = \dfrac{\frac{3mg}{4}(a)^2}{2a} + mga + \dfrac{1}{2}mv^2$ | M1, A1 | Energy equation. Must contain 2 EPE terms, GPE and KE. |
| $v = \dfrac{\sqrt{ag}}{2}$ | A1 | |
---
# Question 6(d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $-\dfrac{a}{3} = \dfrac{2a}{3}\cos(\omega t)$ or $\dfrac{a}{3} = \dfrac{2a}{3}\sin(\omega t)$, $\dfrac{T}{4} = \pi\sqrt{\dfrac{a}{3g}}$ | M1, A1ft | Use $x = "a"\cos(\omega t)$ or $x="a"\sin(\omega t)$. Amplitude consistent with their $AP$. A1: correct equation in $\omega$. |
| $\cos(\omega t) = -\dfrac{1}{2}$ | | |
| $t\sqrt{\dfrac{3g}{4a}} = \dfrac{2\pi}{3}$ | DM1, A1 | Solve to find $t$. Must have full method for complete time. |
| $t = \dfrac{2\pi}{3}\sqrt{\dfrac{4a}{3g}}$ | | Any correct equivalent. Must come from correct working throughout; condone sign error in (b). |6. A light elastic string has natural length $a$ and modulus of elasticity $\frac { 3 } { 4 } \mathrm { mg }$. A particle $P$ of mass $m$ is attached to one end of the string. The other end of the string is attached to a fixed point $A$. Particle $P$ hangs freely in equilibrium at the point $O$, vertically below $A$.
\begin{enumerate}[label=(\alph*)]
\item Find the distance $O A$.
The particle $P$ is now pulled vertically down to a point $B$, where $A B = 3 a$, and released from rest.
\item Show that, throughout the subsequent motion, $P$ performs only simple harmonic motion, justifying your answer.
The point $C$ is vertically below $A$, where $A C = 2 a$.\\
Find, in terms of $a$ and $g$,
\item the speed of $P$ at the instant that it passes through $C$,
\item the time taken for $P$ to move directly from $B$ to $C$.
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VIXV SIHIANI III IM IONOO & VIAV SIHI NI JYHAM ION OO & VI4V SIHI NI JLIYM ION OO \\
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\hfill \mbox{\textit{Edexcel M3 2020 Q6 [15]}}