FM1 3.1 - Hooke's law for strings and springs

You already know

Forces are vectors. A string pulls along its length, and a particle in equilibrium has zero resultant force. Draw the forces on the particle, then resolve them in suitable directions; do not assume a tension equals the weight without checking the geometry.

Natural length, extension and modulus

The natural length $l$ is the length of an unstretched elastic string or spring. When it is stretched straight to length $L>l$, its extension is $x=L-l$. Its modulus of elasticity $\lambda$ is the positive constant in Hooke's law, measured in newtons. It is not a spring stiffness: stiffness has units $\mathrm{N\,m^{-1}}$.

Hooke's law for an elastic string

For a taut ideal elastic string of natural length $l>0$, modulus $\lambda>0$ and extension $x>0$,

At natural length the tension is zero. If the endpoint separation is less than $l$, the string is slack and $T=0$, rather than a negative tension. Use consistent length units and assume the model obeys Hooke's law over the stated range.

Not in the formula booklet: learn it.

Worked example 1: a taut string

Question. A light elastic string has natural length $2\,\mathrm{m}$ and modulus $60\,\mathrm{N}$. Its endpoints are held $2.3\,\mathrm{m}$ apart. Find its tension.

Plan. Subtract the natural length from the endpoint separation, then apply Hooke's law.

Why this step? The numerator contains the extension $0.3$, while the denominator remains the natural length $2$. They are different quantities.

Answer. The tension is $9\,\mathrm{N}$, pulling each endpoint towards the other.

Check. The extension is $0.3/2=0.15$ of the natural length, so $T=0.15(60)=9\,\mathrm{N}$.

Worked example 2: a string cannot push

Question. A light elastic string has natural length $2\,\mathrm{m}$ and modulus $60\,\mathrm{N}$. Its endpoints are held $1.8\,\mathrm{m}$ apart. Find its tension.

Plan. Compare endpoint separation with natural length before applying Hooke's law.

Why this step? The negative difference $1.8-2$ does not describe a compressed string pushing its ends apart. The string simply loses tension and need not remain straight.

Answer. The tension is $0\,\mathrm{N}$.

Check. A string cannot exert the compressive force that a negative value of $\lambda(L-l)/l$ would suggest.

Watch out

Wrong for Worked example 1: $T=60(2.3)/2$. The numerator needs $2.3-2$, not the stretched length. Wrong for Worked example 2: a tension of $-6\,\mathrm{N}$. A slack string has zero tension, not negative tension.

Stiffness and two stretched states

Writing

gives $T=k(L-l)$, with stiffness $k$ in $\mathrm{N\,m^{-1}}$. For two taut states of the same ideal string,

After finding $k$, use $l=L_1-T_1/k$, then $\lambda=kl$. These equations require the same unchanged string and a valid Hooke's-law model.

Not in the formula booklet: learn it.

Worked example 3: two tension measurements

Question. A light elastic string has tension $6\,\mathrm{N}$ when its length is $2.1\,\mathrm{m}$, and tension $24\,\mathrm{N}$ when its length is $2.4\,\mathrm{m}$. Find its natural length and modulus.

Plan. Subtract the two Hooke's-law equations to find stiffness, then recover natural length and modulus.

Why this step? The change in length is also the change in extension, because both measurements use the same natural length.

Answer. The natural length is $2\,\mathrm{m}$, and the modulus is $120\,\mathrm{N}$.

Check. The extensions are $0.1$ and $0.4\,\mathrm{m}$. Hooke's law gives $120(0.1)/2=6\,\mathrm{N}$ and $120(0.4)/2=24\,\mathrm{N}$.

Worked example 4: a changed first reading

Question. A light elastic string has tension $12\,\mathrm{N}$ when its length is $2.1\,\mathrm{m}$, and tension $24\,\mathrm{N}$ when its length is $2.4\,\mathrm{m}$. Find its natural length and modulus.

Plan. Find the new slope of tension against length and use it in either original equation.

Why this step? The smaller change in tension over the same length interval gives a smaller stiffness. A natural length must still be found before converting stiffness to modulus.

Answer. The natural length is $1.8\,\mathrm{m}$, and the modulus is $72\,\mathrm{N}$.

Check. The extensions are $0.3$ and $0.6\,\mathrm{m}$; $72(0.3)/1.8=12\,\mathrm{N}$ and $72(0.6)/1.8=24\,\mathrm{N}$. Both lengths exceed the natural length, as required.

Watch out

Wrong: the gradient $\Delta T/\Delta L$ is the modulus. Its units are $\mathrm{N\,m^{-1}}$, so it is stiffness $k=\lambda/l$. Multiply by the natural length to obtain the modulus in newtons.

Compression and restoring force

A spring of natural length $l$, compressed to length $L<l$, has compression $c=l-L>0$. Its restoring force pushes outwards; a stretched spring pulls inwards. “Restoring” describes the force's direction towards natural length, not necessarily the direction in which the attached particle is moving.

Hooke's law for an ideal spring

With natural length $l>0$, modulus $\lambda>0$ and stiffness $k=\lambda/l$,

If $\delta=L-l$ is signed deformation, the force on the right end of a spring fixed at its left end is

This assumes an ideal light spring obeying Hooke's law over the stated extensions and compressions.

Not in the formula booklet: learn it.