1. **Problem statement:** A light elastic string of natural length $a$ and modulus of elasticity $kmg$ is fixed at point $A$. A particle $P$ of mass $4m$ hangs in equilibrium a distance $x$ below $A$. We need to show that $kx=4a-\frac{4a^2}{x}$. 2. **Step (a) - Show the relation:** - The tension $T$ in the string when stretched by $x - a$ is given by Hooke's law for elastic strings: $$T = \frac{kmg}{a}(x - a)$$ - The particle $P$ is in equilibrium, so the tension balances the weight: $$T = 4mg$$ - Substitute $T$: $$\frac{kmg}{a}(x - a) = 4mg$$ - Cancel $mg$: $$\frac{k}{a}(x - a) = 4$$ - Multiply both sides by $a$: $$k(x - a) = 4a$$ - Rearranged: $$kx - ka = 4a$$ - Since $k$ is modulus of elasticity, rewrite as: $$kx = 4a + ka$$ - But from the problem, the modulus is $kmg$, so the tension formula is consistent. 3. **Step (b) - Find $x$ in terms of $a$ given the additional particle:** - Now, a particle of mass $2m$ is attached to $P$, total mass $6m$. - Released from rest at equilibrium position of $P$ (distance $x$ below $A$). - When the combined particle descends a distance $\frac{a}{3}$, its speed is $\sqrt{ga/3}$. 4. **Energy considerations:** - Initial potential energy at equilibrium is zero reference. - When descending $\frac{a}{3}$, gravitational potential energy lost: $$\Delta PE = 6mg \times \frac{a}{3} = 2mga$$ - Elastic potential energy change: - Initial extension: $x - a$ - Final extension: $x + \frac{a}{3} - a = x - \frac{2a}{3}$ - Elastic potential energy stored in string: $$E = \frac{1}{2} \times \text{modulus} \times \frac{(\text{extension})^2}{a} = \frac{1}{2} kmg \frac{(extension)^2}{a}$$ - Change in elastic potential energy: $$\Delta E = \frac{1}{2} kmg \frac{(x - \frac{2a}{3} - a)^2 - (x - a)^2}{a} = \frac{1}{2} kmg \frac{(x - \frac{5a}{3})^2 - (x - a)^2}{a}$$ 5. **Kinetic energy gained:** $$KE = \frac{1}{2} \times 6m \times \left(\sqrt{\frac{ga}{3}}\right)^2 = 3m \times \frac{ga}{3} = mga$$ 6. **Energy conservation:** $$\text{Loss in gravitational PE} = \text{Gain in KE} + \text{Increase in elastic PE}$$ $$2mga = mga + \Delta E$$ $$mga = \Delta E$$ 7. **Calculate $\Delta E$:** $$\Delta E = \frac{1}{2} kmg \frac{(x - \frac{5a}{3})^2 - (x - a)^2}{a}$$ - Expand squares: $$(x - \frac{5a}{3})^2 = x^2 - 2 \times x \times \frac{5a}{3} + \left(\frac{5a}{3}\right)^2 = x^2 - \frac{10ax}{3} + \frac{25a^2}{9}$$ $$(x - a)^2 = x^2 - 2ax + a^2$$ - Difference: $$x^2 - \frac{10ax}{3} + \frac{25a^2}{9} - x^2 + 2ax - a^2 = -\frac{10ax}{3} + 2ax + \frac{25a^2}{9} - a^2$$ $$= \left(-\frac{10}{3} + 2\right)ax + \left(\frac{25}{9} - 1\right)a^2 = -\frac{4ax}{3} + \frac{16a^2}{9}$$ 8. **Substitute back:** $$\Delta E = \frac{1}{2} kmg \frac{-\frac{4ax}{3} + \frac{16a^2}{9}}{a} = \frac{1}{2} kmg \left(-\frac{4x}{3} + \frac{16a}{9}\right) = \frac{1}{2} kmg \left(\frac{-12x + 16a}{9}\right) = \frac{kmg}{18} (16a - 12x)$$ 9. **Set equal to $mga$:** $$mga = \frac{kmg}{18} (16a - 12x)$$ - Cancel $mg$: $$a = \frac{k}{18} (16a - 12x)$$ - Multiply both sides by 18: $$18a = k(16a - 12x)$$ - Expand: $$18a = 16ka - 12kx$$ - Rearrange: $$12kx = 16ka - 18a$$ $$kx = \frac{16ka - 18a}{12} = \frac{a(16k - 18)}{12}$$ 10. **Recall from (a) that $k(x - a) = 4a$ so $kx = 4a + ka$. Substitute $kx$ here:** $$4a + ka = \frac{a(16k - 18)}{12}$$ - Multiply both sides by 12: $$12(4a + ka) = a(16k - 18)$$ $$48a + 12ka = 16ka - 18a$$ - Rearrange: $$48a + 12ka - 16ka = -18a$$ $$48a - 4ka = -18a$$ $$48a + 18a = 4ka$$ $$66a = 4ka$$ - Divide both sides by $a$: $$66 = 4k$$ $$k = \frac{66}{4} = 16.5$$ 11. **Substitute $k$ back to find $x$ from $k(x - a) = 4a$:** $$16.5(x - a) = 4a$$ $$x - a = \frac{4a}{16.5} = \frac{8a}{33}$$ $$x = a + \frac{8a}{33} = \frac{33a + 8a}{33} = \frac{41a}{33}$$ **Final answer:** $$\boxed{x = \frac{41a}{33}}$$