A student investigates how a spring stretches under different loads. She clamps the spring vertically with its top end fixed, and fixes a metre rule vertically alongside the spring so that she can read off its length. (a) State two other pieces of apparatus the student needs for…

Answer: ((a)) A set of slotted masses (with a hanger) to provide each known load, and a set square held against the metre rule (or a fixed pointer/reference mark) so that each length reading is taken at eye level, avoiding parallax error. ((b)) 2.5 cm, 5.0 cm, 7.5 cm, 10.0 cm (for loads 1.0 N, 2.0 N, 3.0 N, 4.0 N respectively). ((c)) The graph is a straight line passing through the origin. This shows that the extension is directly proportional to the load over this range. Equal increases in load produce equal increases in extension.. Part (a): Apparatus and method In addition to the spring, clamp stand and metre rule, the student needs: - A set of slotted masses and a hanger, to provide a known load (weight) for the spring to support. - A set square (or a fixed pointer…

IGCSE Physics 0625: Load-Extension Graph for a Spring

Load, $F$ / N	0	1.0	2.0	3.0	4.0
Length of spring, $l$ / cm	12.0	14.5	17.0	19.5	22.0

Load,

F

/ N

1.0

2.0

3.0

4.0

Length of spring,

l

/ cm

12.0

14.5

17.0

19.5

22.0

Part (a): Apparatus and method

In addition to the spring, clamp stand and metre rule, the student needs:

A set of slotted masses and a hanger, to provide a known load (weight) for the spring to support.
A set square (or a fixed pointer/reference mark), held against the metre rule at the bottom of the spring, so that each length reading is taken at eye level. This avoids a parallax error, which would otherwise happen if the ruler is read at an angle.

For each load, she reads the length of the spring from the metre rule using this method, and repeats this for a series of increasing loads.

Part (b): Calculating the extension

The extension is the increase in length from the spring’s original (unstretched) length, which is $12.0\text{ cm}$ (when the load is $0\text{ N}$ ):

$\text{extension} = \text{stretched length} - \text{original length}$

Load, $F$ / N	Length, $l$ / cm	Extension $= l - 12.0$ / cm
$1.0$	$14.5$	$2.5$
$2.0$	$17.0$	$5.0$
$3.0$	$19.5$	$7.5$
$4.0$	$22.0$	$10.0$

Part (c): Shape and meaning of the graph

Plotting extension against load using these values gives points $(1.0, 2.5)$ , $(2.0, 5.0)$ , $(3.0, 7.5)$ , $(4.0, 10.0)$ , together with $(0, 0)$ .

Since each $1.0\text{ N}$ increase in load produces exactly the same $2.5\text{ cm}$ increase in extension, these points all lie on a single straight line that passes through the origin.

A straight line through the origin shows that the extension is directly proportional to the load: doubling the load doubles the extension, and equal increases in load always produce equal increases in extension, over this range of results.

Final answers

(a) Slotted masses (and hanger) to provide the load, and a set square/pointer to take each reading at eye level and avoid parallax error
(b) Extensions: $2.5\text{ cm}$ , $5.0\text{ cm}$ , $7.5\text{ cm}$ , $10.0\text{ cm}$
(c) Straight line through the origin. Extension is directly proportional to load