BITSAT 2012 Paper

Step 1: Calculation of spring constant of wireYoung's modulus of wire, $Y = \frac{\frac{\Delta F}{A}}{\frac{\Delta L}{L}} = \frac{\Delta F}{\Delta L} \times \frac{L}{A}$Due to elasticity of the wire, it behaves as equivalent spring $\left(K_{\text{spring}} = \frac{F}{\Delta x}\right)$. Hence, Spring constant for Wire, $K_{W} = \frac{\Delta F}{\Delta L} = Y \frac{A}{L}$Step 2: Calculation of Equivalent spring constantLet a force $F$ be applied to the end of the spring.The wire will extend.Total extension $\Delta S = x_{\text{wire}} + x_{\text{spring}} = \frac{F_{w}}{K_{w}} + \frac{F_{S}}{K}$But $F_S = F_w$ (By Newton's, third Law, Both will apply equal and opposite forces at their joint A)$\therefore \Delta S = F_{w} \left( \frac{1}{K_{w}} + \frac{1}{K} \right)$ So, Equivalent spring constant $(K_{eq})$ is given by:$K_{eq} = \frac{F_{eq}}{X_{eq}} = \frac{F_{w}}{\Delta S} = \frac{1}{\frac{1}{K_{w}} + \frac{1}{K}}$$K_{eq} = \frac{K K_{w}}{K_{w} + K} = \frac{K Y \frac{A}{L}}{Y \frac{A}{L} + K} = \frac{KYA}{YA + KL}$Step 3: Calculation of Time periodAngular frequency $\omega$ for spring block system is given by:$\omega = \sqrt{\frac{K_{eq}}{m}} = \sqrt{\frac{KYA}{m (YA + KL)}}$$\therefore \text{Time period, } T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{(YA + KL) m}{KYA}}$