Page 98 - Physics - XI

Y Scale used on:
X-axis, 1 cm = 50 g wt
Y-axis, 1 cm = 0.5 cm

l B A
3
Extension (l) in cm l l 2 D G θ C

1

θ E F
0 50 100 150 X
Load (w) in g wt

Fig. 2.2: Load-extension graph for helical spring

Therefore, 1 = w = slope of straight line
k l
AC BD 1
tanθ = = =
GC EF k

EF
So, k = = ______ = ______ g wt/cm = ______ N/m.
BD
Convert g wt into newton and cm into metre using the relation.

1 g wt = 10 kg wt = 10 × 9.8 N
–3
–3
and 1 cm = 10 m
–2
Result

The spring constant of the given helical spring by load-extension graph is ______ Nm .
–1
Precautions

1. The helical spring should be suspended freely from a rigid support.
2. The spring should not be loaded beyond the elastic limit of the spring.

3. After every addition or removal of the slotted weights, wait for sometime to take the reading of the
pointer because the spring takes some time to attain its new equilibrium position.
4. Adjust the position of the vertical scale so that the pointer moves in front of the graduations freely
without touching the scale.

Sources of Error
1. Support may not be perfectly rigid so some error may creep in due to the yield of the support.

2. The slotted weights may not have correct weight.

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