Further Mechanics: Elastic Strings, Springs and Elastic Energy
Within A-level Further Mechanics, Hooke's law states that the tension in a stretched elastic string, or the tension or thrust in a stretched or compressed spring, is T = lambda*x/L, where lambda is the modulus of elasticity, x is the extension (or compression) from the natural length, and L is the natural length. A string can only pull, so it exerts no force whenever it is not longer than its natural length; a spring can both pull when stretched and push (a thrust) when compressed. The elastic potential energy (EPE) stored in a stretched or compressed string or spring is EPE = lambda*x^2/(2L), and this energy can be exchanged with kinetic and gravitational potential energy in problems where a particle moves under gravity while attached to an elastic string or spring.
Before you start
Make sure you're comfortable with these topics first:
Method
- Identify the natural length L and the modulus of elasticity lambda from the question; if lambda is not given directly, find it by substituting a known tension and extension into Hooke's law and rearranging.
- Apply Hooke's law, T = lambda*x/L, for the tension in a stretched string or spring, where x is the current length minus the natural length.
- Remember that a string can only pull: it exerts tension only while x>0 (stretched), and exerts no force at all (it is slack) whenever its length is at most its natural length. A spring can also be compressed, exerting a thrust of the same magnitude lambda*x/L outward.
- Use the elastic potential energy formula EPE = lambda*x^2/(2L) for a stretched string or spring, or for a compressed spring, noting EPE depends on x^2 so is the same for an extension x as for a compression of the same size x.
- For a particle moving under gravity while attached to an elastic string or spring, use conservation of energy: total initial energy (KE + GPE + EPE) equals total energy at the point of interest, since gravity and the elastic force are the only forces doing work and both are conservative.
- Where a particle attached to an elastic string is momentarily at rest (for example at its lowest or highest point), its kinetic energy there is 0; use this to find an unknown extension, height or speed elsewhere in the motion.
- Check whether the string is taut throughout the motion described: Hooke's law and the EPE formula apply only once x>0; while the string is slack, treat that stage as free motion under gravity alone, with no tension and no EPE term.
Worked example
One end of a light elastic string of natural length 1.5 m and modulus of elasticity 9.8 N is attached to a fixed point O. A particle of mass 0.5 kg is attached to the other end and is held with the string just taut (at its natural length), then released from rest and falls vertically. Find the maximum extension of the string. (Take g = 9.8 m/s^2.)
- The particle starts with the string at its natural length (x=0) and is momentarily at rest again at the point of maximum extension x_max, having also started from rest, so the loss in gravitational potential energy over the fall equals the elastic potential energy gained: mg*x_max = lambda*x_max^2/(2L).
- Substitute the given values: 0.5(9.8)*x_max = 9.8*x_max^2/(2x1.5), i.e. 4.9*x_max = 9.8*x_max^2/3.
- Multiply both sides by 3: 14.7*x_max = 9.8*x_max^2.
- Since x_max is not zero, divide both sides by x_max: 14.7 = 9.8*x_max.
- Solve: x_max = 14.7/9.8 = 1.5 m.
Practice questions
Try each question, then tap to reveal the answer.
Q1State Hooke's law for an elastic string, defining every symbol used.Show answer
Answer: T = lambda*x/L, where T is the tension, lambda is the modulus of elasticity, x is the extension (current length minus natural length), and L is the natural length.
Q2State the formula for the elastic potential energy stored in a stretched elastic string or spring.Show answer
Answer: EPE = lambda*x^2/(2L), using the same symbols as Hooke's law.
Q3A spring of natural length 0.5 m and modulus of elasticity 15 N is compressed to a length of 0.4 m. Find the thrust in the spring.Show answer
Answer: Compression x = 0.5-0.4 = 0.1 m; thrust = lambda*x/L = 15(0.1)/0.5 = 3 N.
Q4A string of natural length 1.2 m has modulus of elasticity 24 N. Find the extension when the tension is 9 N.Show answer
Answer: 9 = 24x/1.2, so 9 = 20x, giving x = 0.45 m.
Q5Explain why a string, unlike a spring, cannot exert a thrust when compressed.Show answer
Answer: A string has no rigidity, so if it is not stretched beyond its natural length it simply goes slack and exerts no force at all, whereas a spring has the structural rigidity to resist being pushed together and so exerts an outward thrust when compressed.
Q6A light elastic string of natural length 0.6 m and modulus of elasticity 12 N is stretched to a length of 0.9 m. Find the elastic potential energy stored.Show answer
Answer: x = 0.9-0.6 = 0.3 m; EPE = 12(0.3)^2/(2x0.6) = 12(0.09)/1.2 = 1.08/1.2 = 0.9 J.
Q7A string of natural length 0.9 m stores 4.5 J of elastic potential energy when it is stretched by 0.3 m. Find its modulus of elasticity.Show answer
Answer: 4.5 = lambda(0.3)^2/(2x0.9) = lambda(0.09)/1.8 = 0.05*lambda, so lambda = 4.5/0.05 = 90 N.
Q8A particle hangs in equilibrium from an elastic string. Explain why the extension of the string at this equilibrium position is directly proportional to the mass of the particle, for a given string.Show answer
Answer: At equilibrium, tension equals weight, so lambda*x/L = mg. Since lambda and L are fixed for a given string, rearranging gives x = (gL/lambda)*m, and gL/lambda is a constant, so x is directly proportional to m.
Exam-style questions
Written in the style of a A Level Further Maths exam paper, with a full mark scheme.
One end of a light elastic string of natural length 1.5 m and modulus of elasticity 24.5 N is attached to a fixed point O. A particle of mass 0.5 kg is attached to the other end, held at O, and released from rest, falling vertically. (a) Show that the maximum extension x of the string satisfies 5x^2 - 3x - 4.5 = 0, and hence find the maximum extension in exact surd form. (b) Find the speed of the particle at the instant the string first becomes taut. (Take g = 9.8 m/s^2.)
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A block of mass 2 kg slides along a smooth horizontal surface at 2 m/s and collides with the free end of a spring of natural length 0.5 m and modulus of elasticity 25 N; the spring's other end is fixed to a wall. (a) Find the maximum compression of the spring. (b) Find the speed of the block at the instant the spring's compression is 0.2 m, as the block rebounds away from the wall.
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