A hollow tube, sealed at one end, has a cross-sectional area A of 24 cm². The tube contains sand so that the total mass M of the tube and sand is 0.23 kg. The tube floats upright in a liquid of density ρ, as illustrated in Fig. 3.1. [Figure 3.1] The depth of the bottom of the tube below the liquid surface is h. The tube is displaced vertically and then released. The variation with time t of the depth h is shown in Fig. 3.2. [Figure 3.2]
- (a)(i)Determine: the amplitude, in metres, of the oscillations[1]
- (a)(ii)the frequency of oscillation of the tube in the liquid[2]
- (a)(iii)the acceleration of the tube when h is a maximum.[2]
- (b)The frequency f of oscillation of the tube is given by the expression f = 1/(2π) √((Aρg)/M) where g is the acceleration of free fall. Calculate the density ρ of the liquid in which the tube is floating.[2]
- (c)The oscillations illustrated in Fig. 3.2 are undamped. In practice, the liquid does cause light damping. On Fig. 3.2, draw a line to show light damping of the oscillations for time t = 0 to time t = 1.4s.[3]
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Mark scheme
Marking points from the official mark scheme (May/June 2019). The mark or mark type is shown in brackets.
- [1] amplitude = 0.020 m
- [1] T = 0.60 s
- [1] f = 1/T
- [1] = 1.7 Hz
- [1] a = (-)ω²x and [ω = 2πf or ω = 2π/T]
- [1] a = (4π² / 0.60²) × 2.0 × 10⁻²
- [1] = 2.2 m s⁻²
- [1] 1.67 = (1 / 2π) × [(24 × 10⁻⁴ × ρ × 9.81)/0.23]^(1/2)
- [1] ρ = 1.1 × 10³ kg m⁻³
- [1] wave starting with a peak at (0,6)
- [1] wave with same period (or slightly greater)
- [1] peak height decreasing successively
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