Given:
Relaxed length $l_0 = 30$ cm $= 0.3$ m.
Force $F = 60$ N.
Condition: The $p=6^{th}$ turn moves to the position previously occupied by the $q=8^{th}$ turn.

Logic:
In a linear spring, the strain is uniform. The position $x$ of any point on the spring changes to $x’ = x(1 + \epsilon)$, where $\epsilon = \frac{\Delta L}{L_0}$.
Let the spring have $N$ total turns. The position of the $n^{th}$ turn in the relaxed state is $x_n = \frac{n}{N} l_0$.

The new position of the $p^{th}$ turn is:

$$ x’_p = x_p(1 + \epsilon) = \frac{p}{N} l_0 (1 + \epsilon) $$

We are given that this equals the old position of the $q^{th}$ turn:

$$ x’_p = x_q \implies \frac{p}{N} l_0 (1 + \epsilon) = \frac{q}{N} l_0 $$ $$ p(1 + \epsilon) = q \implies 1 + \epsilon = \frac{q}{p} \implies \epsilon = \frac{q-p}{p} $$

Substituting values ($p=6, q=8$):

$$ \epsilon = \frac{8-6}{6} = \frac{2}{6} = \frac{1}{3} $$

From Hooke’s Law, $F = k \Delta L = k (\epsilon l_0)$. Thus, $k = \frac{F}{\epsilon l_0}$.

$$ k = \frac{60}{(1/3)(0.3)} = \frac{60}{0.1} = 600 \text{ N/m} $$

Given:
End A shifts by $\Delta l_A = 5$ cm (Left/Outward).
End B shifts by $\Delta l_B = 25$ cm (Right/Outward).
We need the shift of a point P initially at distance $x = l_0/n$ from A ($n=3$).

Logic:
Since the spring is linear, the displacement $u(x)$ varies linearly with the initial position $x$.
Let $x=0$ be end A and $x=l_0$ be end B.

• Displacement of A ($x=0$): $u(0) = -\Delta l_A$ (Assuming x-axis points A to B, A moves left).
• Displacement of B ($x=l_0$): $u(l_0) = +\Delta l_B$.

Displacement at $x = l_0/n$ is given by linear interpolation:

$$ u(x) = u(0) + \frac{u(l_0) – u(0)}{l_0} \cdot x $$ $$ \Delta x_P = -\Delta l_A + \frac{\Delta l_B – (-\Delta l_A)}{l_0} \cdot \frac{l_0}{n} $$ $$ \Delta x_P = -\Delta l_A + \frac{\Delta l_B + \Delta l_A}{n} $$ $$ \Delta x_P = \frac{-n \Delta l_A + \Delta l_B + \Delta l_A}{n} = \frac{\Delta l_B – (n-1)\Delta l_A}{n} $$

Substituting values ($n=3, \Delta l_A = 5, \Delta l_B = 25$):

$$ \Delta x_P = \frac{25 – (2)(5)}{3} = \frac{15}{3} = 5 \text{ cm} $$