A sonometer wire is vibrating in resonance with a tuning fork. Keeping the tension applied the same, the length of the wire is doubled. Under what condition, would the tuning fork still be in resonance with the wire?

Hint: Use the formula of the frequency of a standing wave.
Wire of twice the length vibrates in its second harmonic. Thus, if the tuning fork resonates at L, it will resonate at 2L. This can be explained below.
Step 1: Find the initial frequency.
The sonometer frequency is given by,
$\mathrm{\nu }=\frac{\mathrm{n}}{2\mathrm{L}}\sqrt{\frac{\mathrm{T}}{\mathrm{m}}}$                                       (n = number of loops)
Now, as it vibrates with length L, we assume, $\mathrm{\nu }={\mathrm{\nu }}_{1}$ and $\mathrm{n}={\mathrm{n}}_{1}$
$\therefore$
Step 2: Find the final frequency.
When the length is doubled, then,

Step 3: Find the no. of loops in the second case.
Dividing Eq. (i) by Eq.(ii), we get,
$\frac{{\mathrm{\nu }}_{1}}{{\mathrm{\nu }}_{2}}=\frac{{\mathrm{n}}_{1}}{{\mathrm{n}}_{2}}×2$
To keep the resonance,
$\frac{{\mathrm{\nu }}_{1}}{{\mathrm{\nu }}_{2}}=1=\frac{{\mathrm{n}}_{1}}{{\mathrm{n}}_{2}}×2$
$⇒$                                ${\mathrm{n}}_{2}=2{\mathrm{n}}_{1}$
Hence, when the wire is doubled the number of loops also doubled to produce the resonance. That is it is the second harmonic.