A simple pendulum consists of a metal sphere suspended from a fixed point by means of a thread, as illustrated in [Figure 3.1]. The sphere of mass 94.0 g is displaced to one side through a horizontal distance of 12.7 cm. The centre of gravity of the sphere rises vertically by 0.90 cm. The sphere is released so that it oscillates. The sphere may be assumed to oscillate with simple harmonic motion.
- (a)State what is meant by simple harmonic motion.[2]
- (b)(i)State the kinetic energy of the sphere when the sphere returns to the displaced position shown in [Figure 3.1].[1]
- (b)(ii)Calculate the total energy E_T of the oscillations.[2]
- (b)(iii)Use your answer in (ii) to show that the angular frequency ω of the oscillations of the pendulum is 3.3 rad s⁻¹.[2]
- (c)The period T of oscillation of the pendulum is given by the expression T = 2π√(L/g) where g is the acceleration of free fall and L is the length of the pendulum. Use data from (b) to determine L.[3]
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Mark scheme
Marking points from the official mark scheme (Oct/Nov 2020). The mark or mark type is shown in brackets.
- [B1] acceleration (directly) proportional to displacement
- [B1] acceleration is in opposite direction to displacement OR acceleration is (directed) towards a fixed point
- [B1] zero
- [C1] ET is maximum potential energy = mgh
- [A1] = 94 x 10⁻³ × 9.81 × 0.90 × 10⁻², = 8.3×10⁻³ J
- [C1] EMAX = ½ mv₀² and v₀ = ωX₀ OR EMAX = ½m(ωX₀)²
- [A1] 8.3 × 10⁻³ = ½ × 94 × 10⁻³ × ω² × (12.7 × 10⁻²)², ...leading to ω = 3.3 rad s⁻¹
- [C1] T = 2π/ω
- [C1] 2π/ 3.3 = 2π × (L / 9.81)½
- [A1] L = 0.90 m
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