Questions on oscillations on strings and springs

Question 1

Explain why a mass hanging on a spring oscillates about the equilibrium position rather than the natural length.

Accepted 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.

Question 2

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.

Accepted 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.

Question 3

A particle of mass 2 kg hangs from a light spring. When it hangs in equilibrium the extension of the spring is 0.05 m. Find the period of small vertical oscillations, taking g = 9.8 m/s².

Accepted Answer

No modulus is given, so use the equilibrium to find the stiffness: λ/l = mg/e = 2 × 9.8/0.05 = 392 N/m. Then ω² = λ/(ml) = 392/2 = 196, which is the same as g/e, so ω = 14 rad/s and T = 2π/14 = 0.449 s. M1 for λ/l = mg/e, A1 for ω² = 196, A1 for the period. The mass cancels: ω² = g/e whatever the mass, so a heavier particle on the same spring stretches it more and oscillates more slowly.

Question 4

A particle of mass m hangs in equilibrium from a light spring of natural length l and modulus λ, with extension e. The particle is then displaced a distance x below the equilibrium position. Show that its equation of motion is mẍ = −(λ/l)x, and hence write down ω².

Accepted Answer

At equilibrium the tension balances the weight: λe/l = mg. Displaced by x below equilibrium the extension is e + x, so the tension is λ(e + x)/l, and taking downwards as positive, mẍ = mg − λ(e + x)/l = mg − λe/l − λx/l. The first two terms cancel by the equilibrium equation, leaving mẍ = −(λ/l)x, which is simple harmonic with ω² = λ/(ml). M1 for the equilibrium equation, M1 for the equation of motion with extension e + x, A1 for the cancellation and ω². The weight cancels only because x is measured from the equilibrium position; measured from the natural length a constant mg would remain in the equation.

Question 5

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.

Accepted 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.

Question 6

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.

Accepted 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.

Question 7

State what changes if the spring in the previous question is replaced by a string of the same natural length and modulus.

Accepted 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.

Question 8

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.

Accepted Answer

Displacing the particle by x stretches one spring by an extra x and shortens the other by x. Both changes of tension act back towards the centre, each of size 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. The two tensions pull opposite ways, but their changes both act towards the centre, so they add rather than cancel.

Question 9

A particle of mass 0.25 kg hangs from a light spring of natural length 0.5 m and modulus 50 N and oscillates vertically. Its displacement below the equilibrium position after t seconds is x = 0.04 sin 20t. Find the tension in the spring when t = 0.1 s, and show that at the highest point of the motion the spring is in compression. Take g = 9.8 m/s².

Accepted Answer

The equilibrium extension is e = mgl/λ = 0.25 × 9.8 × 0.5/50 = 0.0245 m. (Check: ω² = λ/(ml) = 50/(0.25 × 0.5) = 400, so ω = 20 as the equation says.) At t = 0.1, x = 0.04 sin 2 = 0.0364 m below equilibrium, so the extension is 0.0245 + 0.0364 = 0.0609 m and the tension is λ × extension/l = 50 × 0.0609/0.5 = 6.09 N. At the highest point x = −0.04, so the extension would be 0.0245 − 0.04 = −0.0155 m: the spring is 0.0155 m shorter than its natural length and exerts a thrust of 50 × 0.0155/0.5 = 1.55 N. B1 for the equilibrium extension, M1 for x at t = 0.1, A1 for 0.0364 m, A1 for the tension 6.09 N, B1 for the negative extension at the top and the compression. The tension is λ/l times the whole extension from the natural length, not times x alone; x is measured from equilibrium, where the spring is already stretched by e.

Question 10

A particle of mass 0.4 kg hangs from a light elastic string of natural length 0.8 m and modulus 39.2 N. It is pulled down 0.05 m below its equilibrium position and released from rest. Show that the string remains taut throughout the motion, and find the greatest tension in it. Take g = 9.8 m/s².

Accepted Answer

Equilibrium: 39.2e/0.8 = 0.4 × 9.8, so 49e = 3.92 and e = 0.08 m. Released from rest 0.05 m below equilibrium, the amplitude is 0.05 m, which is less than 0.08 m, so the particle never rises to the natural length and the string stays taut: the motion is simple harmonic throughout. The greatest tension is at the lowest point, where the extension is 0.08 + 0.05 = 0.13 m: T = 39.2 × 0.13/0.8 = 6.37 N. M1 for equating tension to weight, A1 for e = 0.08 m, B1 for comparing the amplitude with the extension and concluding the string stays taut, M1 for the tension at extension 0.13 m, A1 for 6.37 N. At the top the extension is 0.03 m and the tension 1.47 N, still positive.

Question 11

A light spring of natural length 0.6 m is to carry a particle of mass 0.3 kg hanging vertically, and the period of its small oscillations is to be 0.5 s. Find the modulus of elasticity the spring must have, and the extension at which the particle then hangs in equilibrium. Take g = 9.8 m/s².

Accepted Answer

T = 2π/ω = 0.5 gives ω = 4π rad/s, so ω² = 16π² = 157.9. Then ω² = λ/(ml) gives λ = ω²ml = 157.9 × 0.3 × 0.6 = 28.4 N. In equilibrium λe/l = mg, so e = mgl/λ = 0.3 × 9.8 × 0.6/28.4 = 0.0621 m, which is also g/ω² = 9.8/157.9. M1 for ω from the period, M1 for λ = ω²ml, A1 for 28.4 N, A1 for the extension.

Question 12

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.

Accepted 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.

Question 13

A particle of mass 0.5 kg hangs in equilibrium from a light elastic string of natural length 1.2 m and modulus 29.4 N. It is then projected vertically downwards with speed u m/s. Find the greatest value of u for which the string never goes slack, and show that with this value of u the greatest tension in the string is twice the weight of the particle. Take g = 9.8 m/s².

Accepted Answer

Equilibrium: 29.4e/1.2 = 0.5 × 9.8, so e = 0.2 m, and ω² = λ/(ml) = 29.4/(0.5 × 1.2) = 49, so ω = 7 rad/s.
Projected from the centre of the oscillation at speed u, the amplitude is a = u/ω, since u is the greatest speed aω. The string stays taut as long as the particle never rises above the natural length, that is a ≤ e: u/7 ≤ 0.2, so u ≤ 1.4 m/s.
With u = 1.4 the amplitude is 0.2 m and the lowest point has extension e + a = 0.4 m, where the tension is 29.4 × 0.4/1.2 = 9.8 N, which is 2 × 0.5 × 9.8 = 2mg.
B1 for e = 0.2 m, B1 for ω = 7, M1 for the condition u/ω ≤ e, A1 for 1.4 m/s, M1 A1 for the tension at extension 2e and the comparison with 2mg. The result is general: at the limiting amplitude a = e the extension runs from 0 to 2e, so the tension runs from 0 to 2mg whatever the string.

Question 14

A particle of mass 2 kg hangs from a light spring of natural length 0.5 m and modulus 100 N. It is pulled down 0.1 m below its equilibrium position and released from rest. By considering kinetic, gravitational and elastic energy, find its speed as it passes through the equilibrium position, and confirm the value using the simple harmonic result for the greatest speed. Take g = 9.8 m/s².

Accepted Answer

Equilibrium extension e = mgl/λ = 2 × 9.8 × 0.5/100 = 0.098 m, so at the release point the extension is 0.198 m.
From release to equilibrium the elastic energy falls from λ(0.198)²/(2l) to λ(0.098)²/(2l): released energy = (100/1)(0.198² − 0.098²) = 2.96 J. The particle rises 0.1 m, gaining 2 × 9.8 × 0.1 = 1.96 J of gravitational energy. So ½ × 2 × v² = 2.96 − 1.96 = 1.00, giving v = 1.0 m/s.
Check: ω² = λ/(ml) = 100/(2 × 0.5) = 100, so ω = 10, amplitude 0.1 m, greatest speed aω = 1.0 m/s at the centre, in agreement.
M1 for elastic energy at both extensions, A1 for 2.96 J, B1 for the gravitational term 1.96 J, M1 for the energy equation with all three terms, A1 for v = 1.0 m/s, B1 for the aω check. Leaving out the gravitational term gives v² = 2.96, and the wrong answer; leaving out the elastic energy at equilibrium (as though the spring were unstretched there) is the other slip, and the two errors do not cancel.

Question 15

A particle of mass 0.5 kg hangs from a light spring of natural length 0.6 m and modulus 30 N, and performs small vertical oscillations. The spring is then cut in half and the particle is hung from one of the halves. Find the period of small oscillations in each case and the ratio of the new period to the old, and explain why the modulus of elasticity of the half spring is not 15 N.

Accepted Answer

Whole spring: ω² = λ/(ml) = 30/(0.5 × 0.6) = 100, so T = 2π/10 = 0.628 s.
The modulus of elasticity is the tension that would double the length of the spring. When the whole spring carries a tension T each half carries the same tension and stretches by half the total, so each half doubles its own length under exactly the tension that doubles the whole: the half has modulus 30 N and natural length 0.3 m. Its stiffness λ/l is 100 N/m, twice the 50 N/m of the whole spring.
Half spring: ω² = 30/(0.5 × 0.3) = 200, so T = 2π/√200 = 0.444 s. Ratio 0.444/0.628 = 1/√2 ≈ 0.707.
M1 A1 for the first period, B1 for the half spring keeping the modulus 30 N with natural length 0.3 m, B1 for the explanation in terms of the tension that doubles the length, M1 A1 for the second period and the ratio. Halving the length doubles the stiffness, and the period goes as 1/√(stiffness), hence the 1/√2. Taking the modulus to be 15 N would give the same stiffness as before and predict no change at all, which is wrong: a shorter spring of the same material is stiffer.