Problem 4 : A spring of spring constant "k", has extended lengths "
x
1
x
1
" and "
x
2
x
2
" corresponding to spring forces 4 N and 5 N. Find the spring length when spring force is 9 N.

Solution : Let the natural length of spring be “x”. According to Hooke’s law, the magnitude of spring force,

F
=
k
Δ
x
F
=
k
Δ
x

where "
Δ
x
Δ
x
" is the extension. Let the spring length for spring force 9N be "
x
3
x
3
". Then,

9
=
k
x
3
−
x
=
k
x
3
−
k
x
9
=
k
x
3
−
x
=
k
x
3
−
k
x

⇒
x
3
=
9
+
k
x
k
⇒
x
3
=
9
+
k
x
k

We need to find expressions of spring constant and natural length in terms of given values. For the given two extensions, we have :

4
=
k
x
1
−
x
4
=
k
x
1
−
x

5
=
k
x
2
−
x
5
=
k
x
2
−
x

Subtracting first equation from second and solving for “k”,

⇒
1
=
k
x
2
−
x
1
⇒
1
=
k
x
2
−
x
1

k
=
1
x
2
−
x
1
k
=
1
x
2
−
x
1

Substituting for “k” in the first equation,

⇒
4
=
1
x
2
−
x
1
X
x
1
−
x
⇒
4
=
1
x
2
−
x
1
X
x
1
−
x

⇒
4
x
2
−
x
1
=
x
1
−
x
⇒
4
x
2
−
x
1
=
x
1
−
x

⇒
4
x
2
−
4
x
1
=
x
1
−
x
⇒
4
x
2
−
4
x
1
=
x
1
−
x

⇒
x
=
5
x
1
−
4
x
2
⇒
x
=
5
x
1
−
4
x
2

We can, now, evaluate the required expression of “
x
3
x
3

⇒
x
3
=
9
+
1
x
2
−
x
1
X
5
x
1
−
4
x
2
1
x
2
−
x
1
⇒
x
3
=
9
+
1
x
2
−
x
1
X
5
x
1
−
4
x
2
1
x
2
−
x
1

⇒
x
3
=
9
x
2
−
x
1
+
5
x
1
−
4
x
2
⇒
x
3
=
9
x
2
−
x
1
+
5
x
1
−
4
x
2

x
3
=
5
x
2
−
4
x
1
x
3
=
5
x
2
−
4
x
1