Standard +0.8 This is a Further Maths SHM problem requiring application of energy relationships in SHM. Students must use KE = ½m(ω²)(a²-x²) and set up an equation from the given energy ratio, then solve for amplitude. It requires understanding of the energy formula and algebraic manipulation, going beyond standard 'find velocity at a point' questions, but follows a clear method once the approach is identified.
The point \(O\) is on the fixed horizontal line \(l\). Points \(A\) and \(B\) on \(l\) are such that \(OA = 0.1\) m and \(OB = 0.5\) m, with \(B\) between \(O\) and \(A\). A particle \(P\) oscillates on \(l\) in simple harmonic motion with centre \(O\). The kinetic energy of \(P\) when it is at \(A\) is twice its kinetic energy when it is at \(B\). Find the amplitude of the motion.
[3]
Find amplitude a from ratio 2 of [½ m] v 2to [½ m] v 2
A B
Answer
Marks
Guidance
a2 = 0.5 – 0.01 = 0.49, a = 0 .7 [m]
A1
(taking ratio ½ loses A1)
3
Answer
Marks
Guidance
Question
Answer
Marks
Question 1:
1 | v 2 = ω2 (a2 – 0.12) and v 2 = ω2 (a2 – 0.52)
A B | B1 | Use v2 = ω2 (a2 – x2) at A and B (may be implied)
a2 – 0.12 = 2 (a2 – 0.52) | M1 | Find amplitude a from ratio 2 of [½ m] v 2to [½ m] v 2
A B
a2 = 0.5 – 0.01 = 0.49, a = 0 .7 [m] | A1 | (taking ratio ½ loses A1)
3
Question | Answer | Marks | Guidance
The point $O$ is on the fixed horizontal line $l$. Points $A$ and $B$ on $l$ are such that $OA = 0.1$ m and $OB = 0.5$ m, with $B$ between $O$ and $A$. A particle $P$ oscillates on $l$ in simple harmonic motion with centre $O$. The kinetic energy of $P$ when it is at $A$ is twice its kinetic energy when it is at $B$. Find the amplitude of the motion.
[3]
\hfill \mbox{\textit{CAIE FP2 2018 Q1 [3]}}