Two wires, of the same diameter and same material have lengths $1.2\text{ m}$ and $2\text{ m}$, vibrate with…
Two wires, of the same diameter and same material have lengths $1.2\text{ m}$ and $2\text{ m}$, vibrate with the same fundamental frequency. If the shorter wire is stretched by a force of $36\text{ kg-wt}$, then the tension in the longer wire is [EAMCET 2015]
$50\text{ kg-wt}$
$200\text{ kg-wt}$
$100\text{ kg-wt}$
$400\text{ kg-wt}$
Solution
Given, $l_1 = 1.2\text{ m}$, $\alpha_1 = \alpha$,
$F_1 = 36\text{ kg-wt}$ and $l_2 = 2\text{ m}$
Let, $\alpha_2 = \alpha \Rightarrow F_2 = F$
$\nu_2 = \nu_1$
As fundamental frequency,
$\nu_1 = \frac{1}{2l_1} \sqrt{\frac{F_1}{\alpha_1}} = \frac{1}{2 \times 1.2} \sqrt{\frac{36}{\alpha}}$ ...(i)
$\nu_2 = \frac{1}{2 \times 2} \sqrt{\frac{F}{\alpha}}$ ...(ii)
On dividing Eq. (i) by Eq. (ii), we get
$\frac{\nu_1}{\nu_2} = \frac{\frac{1}{2.4} \sqrt{\frac{36}{\alpha}}}{\frac{1}{4} \sqrt{\frac{F}{\alpha}}} = 1$ (Given)
$\Rightarrow \frac{1}{2.4} \sqrt{\frac{36}{\alpha}} = \frac{1}{4} \sqrt{\frac{F}{\alpha}} \Rightarrow \frac{24}{2.4} = \sqrt{F}$
$\therefore F = 100\text{ kg-wt}$