A U-shaped tube contains some liquid. The liquid column in each half of the tube has length L, as shown in [Figure 3.1]. The liquid columns are displaced vertically. The liquid then oscillates in the tube. The liquid levels are displaced from the equilibrium positions as shown in [Figure 3.2]. The acceleration a of the liquid in the tube is related to the displacement x by the expression a = -(g/L)x where g is the acceleration of free fall.
- (a)Explain how the expression shows that the liquid in the tube is undergoing simple harmonic motion.[3]
- (b)The length L of each liquid column is 18 cm. Determine the period T of the oscillations.[3]
- (c)The oscillations of the liquid in the tube are damped. In any one complete cycle of the oscillations, the amplitude decreases by 6.0% of its value at the beginning of the oscillation. Determine the ratio energy of oscillations after 3 cycles / initial energy of oscillations[3]
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Mark scheme
Marking points from the official mark scheme (May/June 2021). The mark or mark type is shown in brackets.
- [1] acceleration in opposite direction to displacement shown by – sign
- [1] g/L is constant
- [1] (so) acceleration is (directly) proportional to displacement
- [1] ω² = g/L
- [1] ω = 2π / Τ OR ω = 2πf and f = 1/T
- [1] T = 0.85 s
- [1] energy ∝ x₀²
- [1] (after 3 cycles,) amplitude = (0.94)³x₀
- [1] ratio final energy / initial energy = 0.83²
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