The piston in the cylinder of a car engine moves in the cylinder with simple harmonic motion. The piston moves between a position of maximum height in the cylinder to a position of minimum height, as illustrated in Fig. 3.1. [Figure 3.1] The distance moved by the piston between the positions shown in Fig. 3.1 is 9.8 cm. The mass of the piston is 640 g. At one particular speed of the engine, the piston completes 2700 oscillations in 1.0 minute. (a) For the oscillations of the piston in the cylinder, determine:
- (a)(i)the amplitude[1]
- (a)(ii)the frequency[1]
- (a)(iii)the maximum speed[2]
- (a)(iv)the speed when the top of the piston is 2.3 cm below its maximum height.[2]
- (b)The acceleration of the piston varies. Determine the resultant force on the piston that gives rise to its maximum acceleration.[3]
This question uses a diagram from the original paper, which isn't shown here. Open the full paper from the past papers archive to see it.
Mark scheme
Marking points from the official mark scheme (May/June 2020). The mark or mark type is shown in brackets.
- [A1] amplitude = 4.9 cm
- [A1] frequency = 2700 / 60 = 45 Hz
- [C1] V₀ = x₀ω and ω = 2πf
- [A1] V₀ = 4.9 x 10⁻² × 2π × 45 = 14 m s⁻¹
- [C1] v = ω (x₀² − x²)½
- [A1] = 2π × 45 × [(4.9 × 10⁻²)² – (2.6 × 10⁻²)²]½ = 12 m s⁻¹
- [C1] F = ma and a₀ = V₀ω or a₀ = x₀ω²
- [C1] F = 0.64 × 13.9 × 2π × 45 or 0.64 × 4.9 × (2π × 45)²
- [A1] = 2500 N
Write your answer and get it marked
On Oracle Prep you can write or photograph your own answer and have it marked by AI against this mark scheme, with feedback on what you missed.
Sign up free →