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Oscillations on strings and springs questions
Hooke's law supplies a restoring force proportional to displacement, which is exactly the simple harmonic condition. The only care needed is where the motion is centred and whether the string stays taut.
7 original questions · 27 marks · the oscillations on strings and springs notes · Further Mechanics 2
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Explain why a mass hanging on a spring oscillates about the equilibrium position rather than the natural length.
Worked answer
At equilibrium the tension already balances the weight. Measuring the displacement from there, the extra tension is λx/l and the weight has cancelled out, leaving a restoring force proportional to x. The weight therefore fixes where the oscillation is centred but has no effect on the motion about that point. B1 B1 for the weight cancelling and for the restoring force proportional to x.A mass of 1 kg hangs on a spring of natural length 0.5 m and modulus 40 N. Find ω and the period of small oscillations.
Worked answer
ω² = λ/(ml) = 40/(1 × 0.5) = 80, so ω = 8.94 rad/s and T = 2π/ω = 0.702 s. M1 for ω² = λ/(ml), A1 for ω, A1 for the period.A mass of 1 kg hangs on a spring of natural length 0.5 m and modulus 40 N. Find the equilibrium extension, taking g = 9.8 m/s².
Worked answer
At equilibrium the tension equals the weight, so 40e/0.5 = 1(9.8). Then 80e = 9.8 and e = 0.1225 m. M1 for equating tension to weight, A1 for 80e = 9.8, A1 for the extension. Note that ω² = λ/(ml) = g/e, a useful check here since 9.8/0.1225 = 80.The mass is raised to the point where the spring has its natural length and released from rest. Find its greatest speed and where it occurs.
Worked answer
Released from rest at 0.1225 m above equilibrium, so the amplitude is 0.1225 m. The greatest speed is aω = 0.1225 × 8.944 = 1.10 m/s, reached at the equilibrium position, where the net force is zero and the acceleration vanishes. B1 for the amplitude, M1 A1 for the greatest speed, B1 for placing it at the equilibrium position.State what changes if the spring in the previous question is replaced by a string of the same natural length and modulus.
Worked answer
A string cannot push, so above the natural length it goes slack and exerts nothing. Released from exactly the natural length, the motion is still simple harmonic throughout, since the particle only just reaches that point. Any greater amplitude and the particle would rise above it as a free body under gravity, so the motion would be simple harmonic for part of each cycle only. B1 for the string going slack above the natural length, B1 for the motion staying simple harmonic in this case, B1 for what a greater amplitude would do.A particle of mass 0.5 kg lies on a smooth horizontal table between two springs, each of natural length 0.4 m and modulus 25 N, attached to fixed points and both taut. Find the period of small oscillations.
Worked answer
Displacing the particle by x stretches one spring by an extra x and shortens the other by x. Both then push it back, each with force 25x/0.4, so the total restoring force is 2 × 25x/0.4 = 125x. Hence 0.5ẍ = −125x, giving ω² = 2λ/(ml) = 250 and ω = 15.8 rad/s. T = 2π/15.8 = 0.397 s. M1 for the total restoring force, A1 for ω² = 250, M1 for T = 2π/ω, A1 for 0.397 s. Two springs make the system stiffer, and the period shortens by a factor of √2. Both springs push the particle back, so the forces add rather than cancel, and that is the step most often got wrong.A particle of mass 0.5 kg hangs at rest from an elastic string of natural length 1 m and modulus 24.5 N attached to a fixed point. It is pulled down 0.5 m below its equilibrium position and released from rest. Find the speed of the particle when the string goes slack, and the greatest height it reaches above the equilibrium position. Take g = 9.8 m/s².
Worked answer
At equilibrium 24.5e/1 = 0.5(9.8), so e = 0.2 m, and ω² = λ/(ml) = 24.5/0.5 = 49, giving ω = 7 rad/s.
While the string is taut the motion is simple harmonic about the equilibrium position with amplitude a = 0.5 m. The string goes slack when the particle is level with the natural length, that is at displacement x = 0.2 m above the centre. Then v² = ω²(a² − x²) = 49(0.25 − 0.04) = 10.29, so v = 3.21 m/s upwards.
Above that point the string exerts nothing, so the particle is a free body under gravity alone and the motion is no longer simple harmonic. It rises a further v²/(2g) = 10.29/19.6 = 0.525 m.
Greatest height above the equilibrium position = 0.2 + 0.525 = 0.725 m.
M1 A1 for the equilibrium extension, B1 for ω = 7, M1 for locating where the string goes slack, M1 A1 for the speed there, M1 A1 for the further rise and the total height. The amplitude 0.5 m exceeds the extension 0.2 m, so the string must go slack, and using v² = ω²(a² − x²) all the way to the top would be wrong. Marks are awarded for saying where the simple harmonic motion stops.
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