A wire of length $L$ and linear density $m$ is stretched by a force $T$ and the frequency is $n_1$. Another…
A wire of length $L$ and linear density $m$ is stretched by a force $T$ and the frequency is $n_1$. Another wire of same material of length $2L$ and same linear density is stretched by a force $9T$ and its frequency is $n_2$. Then, the value of $(n_2/n_1)$ is [EAMCET 2013]
$4 : 1$
$1 : 3$
$3 : 2$
$1 : 2$
Solution
We know that, the frequency, $n = \frac{1}{2l}\sqrt{\frac{T}{\alpha}}$
Here, $\alpha$ is a constant, so $n \propto \frac{\sqrt{T}}{l}$
Now, we can calculate the value of $n_2 / n_1$ as
$\frac{n_2}{n_1} = \frac{l_1}{l_2} \times \sqrt{\frac{T_2}{T_1}} = \frac{1}{2} \times \sqrt{\frac{9}{1}} = \frac{\sqrt{9}}{2} = \frac{3}{2}$
Hence, the ratio of $\frac{n_2}{n_1}$ is $\frac{3}{2}$.