Edexcel M3 — Question 2 7 marks

Exam BoardEdexcel
ModuleM3 (Mechanics 3)
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSimple Harmonic Motion
TypeFind amplitude of SHM
DifficultyStandard +0.3 This is a standard SHM verification question requiring differentiation twice to show acceleration is proportional to negative displacement (routine M3 technique), followed by straightforward application of initial conditions and the amplitude formula √(p²+q²). The question is slightly above average difficulty due to the algebraic manipulation required, but follows a well-practiced template with no novel insight needed.
Spec4.10f Simple harmonic motion: x'' = -omega^2 x

A particle moves along a straight line in such a way that its displacement \(x\) m from a fixed point \(O\) on the line, at time \(t\) seconds after it leaves \(O\), is given by \(x = p \sin \omega t + q \cos \omega t\) where \(p\), \(q\) and \(\omega\) are constants.
  1. Show that the motion of the particle is simple harmonic. [5 marks]
  2. If the particle leaves \(O\) with speed 15 ms\(^{-1}\), and \(\omega = 3\), find the amplitude of the motion. [2 marks]

AnswerMarks
(a) \(x = p\sin\omega t + q\cos\omega t\)M1 A1
\(\dot{x} = p\omega\cos\omega t - q\omega\sin\omega t\)M1 A1
\(\ddot{x} = -p\omega^2\sin\omega t - q\omega^2\cos\omega t = -\omega^2(p\sin\omega t + q\cos\omega t) = -\omega^2 x\)M1 A1
Acceleration proportional to displacement and directed towards \(O\), so SHMB1
(b) Maximum speed at \(O\), so \(15 = a\omega = 3a\)M1 A1
\(a = 5\) mM1 A1
Total: 7 marks
(a) $x = p\sin\omega t + q\cos\omega t$ | M1 A1 |

$\dot{x} = p\omega\cos\omega t - q\omega\sin\omega t$ | M1 A1 |

$\ddot{x} = -p\omega^2\sin\omega t - q\omega^2\cos\omega t = -\omega^2(p\sin\omega t + q\cos\omega t) = -\omega^2 x$ | M1 A1 |

Acceleration proportional to displacement and directed towards $O$, so SHM | B1 |

(b) Maximum speed at $O$, so $15 = a\omega = 3a$ | M1 A1 |

$a = 5$ m | M1 A1 |

| **Total: 7 marks** |
A particle moves along a straight line in such a way that its displacement $x$ m from a fixed point $O$ on the line, at time $t$ seconds after it leaves $O$, is given by $x = p \sin \omega t + q \cos \omega t$ where $p$, $q$ and $\omega$ are constants.

\begin{enumerate}[label=(\alph*)]
\item Show that the motion of the particle is simple harmonic. [5 marks]
\item If the particle leaves $O$ with speed 15 ms$^{-1}$, and $\omega = 3$, find the amplitude of the motion. [2 marks]
\end{enumerate}

\hfill \mbox{\textit{Edexcel M3  Q2 [7]}}