A sonometer wire vibrating in two loops is in resonance with a...

The frequency of a vibrating string is given by $f = \dfrac{p}{2L} \sqrt{\dfrac{T}{\mu}}$, where $p$ is the number of loops, $L$ is the length, $T$ is the tension, and $\mu$ is the linear mass density.

Let the initial frequency be $f_1$. Given $p_1 = 2$, we have:

$f_1 = \dfrac{2}{2L_1} \sqrt{\dfrac{T_1}{\mu}}$

For the final state, the tension is increased to $16$ times the original tension ($T_2 = 16T_1$) and the length is doubled ($L_2 = 2L_1$). The frequency $f_2$ must be equal to $f_1$ for the wire to remain in resonance with the same tuning fork.

$f_2 = \dfrac{p_2}{2L_2} \sqrt{\dfrac{T_2}{\mu}}$

Equating $f_1$ and $f_2$:

$\dfrac{2}{2L_1} \sqrt{\dfrac{T_1}{\mu}} = \dfrac{p_2}{2(2L_1)} \sqrt{\dfrac{16T_1}{\mu}}$

$\dfrac{1}{L_1} \sqrt{\dfrac{T_1}{\mu}} = \dfrac{p_2 \times 4}{4L_1} \sqrt{\dfrac{T_1}{\mu}}$

$\dfrac{1}{L_1} = \dfrac{p_2}{L_1}$

$p_2 = 1$

Therefore, the wire will vibrate in one loop.

Answer: one loop