1.4 - Elasticity

Elasticity and Spring Extension

Elasticity is a material’s ability to regain its original shape and size when the applied force is removed.
An applied force can change an object’s shape and size. An elastic exercise cord returns to its original length after stretching.
Spring extension is the increase in length caused by a stretching force:

x = l - l_{0}

$l_{0}$ is original length; $l$ is stretched length. Extension $x$ is the change in length, rather than the final length.

Apparatus: spring at least 10 cm long; five 10 g, five 20 g and five 50 g slotted weights; half-metre rule; two retort stands with clamps.
Materials: pin, plasticine and thread.

Suspend the spring next to the rule; align the rule’s zero mark with the spring’s upper end. Attach the pin using plasticine as the lower-end pointer.
Record the initial pin position to obtain original length $l_{0}$ .
Increase the applied force using the supplied slotted weights and measure the stretched length and extension.
Record mass, applied force, stretched length and extension; plot extension $x$ against force $F$ .

Precaution: keep extension within half the original length to avoid overstretching and allow the spring to regain its length.
Result: the $x$ against $F$ graph is a straight line through the origin.
Conclusion: spring extension is directly proportional to applied force within the elastic limit.

Hooke’s law: spring extension is proportional to the applied force provided the spring’s elastic limit is not exceeded.

x \propto F, F = k x, k = \frac{F}{x}

$k$ is the spring constant. With $F$ in N and $x$ in m, its unit is $N m^{- 1}$ .
On a force $F$ against extension $x$ graph, $k$ equals the gradient. Keep the graph’s axes in the stated order.

Elastic potential energy $E_{P}$ equals the work done to stretch the spring and the area under the $F$ against $x$ graph.
Within Hooke’s law, force rises linearly from 0 to $F$ , so average force is $(0 + F) /2$ . The graph’s area is a right-angled triangle:

E_{P} = \frac{0 + F}{2} x = \frac{1}{2} F x

E_{P} = \frac{1}{2} k x^{2}

Use $F$ in N and $x$ in m to obtain energy in joules, J.
Comparing spring graphs: at extension 0.20 m, spring R requires 8 N, while spring S requires 16 N. From $k = F / x$ , their spring constants are $40 N m^{- 1}$ and $80 N m^{- 1}$ respectively.
From $E_{P} = \frac{1}{2} F x$ , R stores 0.80 J and S stores 1.60 J at that extension. S has the steeper graph, is stiffer and stores more elastic potential energy at the same extension.

Compare springs by applying equal loads and observing their extensions. The spring with the smaller extension has the larger $k$ .
Compare one characteristic at a time; the spring pairs comparing thickness, diameter and length use the same material.

Factor	Effect on $k$ , with other characteristics unchanged
Material	$k$ varies with the material.
Spring length	Shorter: larger $k$ ; longer: smaller $k$ .
Spring diameter	Smaller coil diameter: larger $k$ ; larger coil diameter: smaller $k$ .
Thickness of spring wire	Thicker wire: larger $k$ ; thinner wire: smaller $k$ .

Distinguish coil diameter from wire thickness: their effects on $k$ differ.

Arrangement of two identical springs	Force and extension
Series	Each spring experiences $F$ and extends by $x$ ; system extension is $x + x = 2 x$ .
Parallel	Each spring experiences $F /2$ and extends by $x /2$ ; system extension is $x /2$ .

In series, add the extensions along the arrangement. In parallel, the total force is shared equally among identical springs and their extensions are equal.

Each spring has original length 50 mm and extends by 6 mm under 12 N:

k = \frac{12}{6} = 2 N m m^{- 1}

Two springs P and Q in series: each tension is 12 N and each extension is 6 mm. System extension is 12 mm; total length is $50 + 50 + 12 = 112 mm$ .
Two springs R and S in parallel: each tension is $12/2 = 6 N$ and each extension is $6/2 = 3 mm$ . System extension is 3 mm; total length is $50 + 3 = 53 mm$ .
Three parallel springs T, U and V in series with W: T, U and V each carry $12/3 = 4 N$ and extend by $4/2 = 2 mm$ . W carries 12 N and extends by 6 mm.
The parallel group extends by 2 mm; the mixed system extends by $2 + 6 = 8 mm$ and has total length $50 + 50 + 8 = 108 mm$ .