Question

One end of a long metallic wire of length L is tied to the ceiling. The other end is tied to massless spring of spring constant K. A mass m hangs freely from the free end of the spring. The area of cross section and the young's modulus of the wire are A and Y respectively. If the mass is slightly pulled down and released, it will oscillated with a time period T equal to.

• A. $2\pi \sqrt{\left(m/K\right)}$
• B. $2\pi \sqrt{m(YA+KL)/\left(YAK\right)}$
• C. $2\pi \sqrt{\left(mYA/KL\right)}$
• D. $2\pi \sqrt{\left(mL/YA\right)}$

Answer 1

The correct option is B $2\pi \sqrt{m(YA+KL)/\left(YAK\right)}$

$\text{Step 1: Calculation of spring constant of wire}$

Young's modulus of wire, $Y=\frac{\Delta F/A}{\Delta L/L}=\frac{\Delta F}{\Delta L}\times \frac{L}{A}$

Due to elasticity of the wire, it behaves as equivalent spring$(K_{spring}=F/\Delta x)$.

Hence, Spring constant for Wire, $K_w=\frac{\Delta F}{\Delta L}=Y\frac{A}{L}$

$\text{Step 2: Calculation of Equivalent spring constant}$

Let a force $F$ be applied to the end of the spring.
The wire will extend.
$\text{Total extension}$ $\Delta S=x_{wire}+x_{spring}=\frac{F_w}{K_w}+\frac{F_S}{K}$

But $F_S=F_w$ $($ By Newton's, third Law, Both will apply equal and opposite forces at their joint A $)$

$\therefore \Delta S=F_w(\frac{1}{K_w}+\frac{1}{K})$

So, $\text{Equivalent spring constant}$ $\left(K_{eq}\right)$ is given by:
$K_{eq}=\frac{F_{eq}}{x_{eq}}=\frac{F_w}{\Delta S}=\frac{1}{\frac{1}{K_w}+\frac{1}{K}}$

$K_{eq}=\frac{\left(K\right)\left(K_w\right)}{K_w+K}=\frac{KY\frac{A}{L}}{Y\frac{A}{L}+K}=\frac{KYA}{YA+KL}$

$\text{Step 3: Calculation of Time period}$

Angular frequency $\omega$ for spring block system is given by:

$\omega =\sqrt{\frac{K_{eq}}{m}}=\sqrt{\frac{KYA}{m(YA+KL)}}$

$\therefore$ Time period, $T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{(YA+KL)m}{KYA}}$

Hence, Option B is correct