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A sonometer wire is vibrating in resonance with a tuning fork. Keeping the tension applied same, the length of the wire is doubled. Under what conditions would the tuning fork still be is resonance with the wire ?
Solution
Initially $f_{1}=\frac{1}{2 L} \sqrt{\frac{\mathrm{T}}{\mu}}=f$ (Frequency of given tuning fork)
Finally, $f_{1}^{\prime}=\frac{1}{2(2 \mathrm{~L})} \sqrt{\frac{\mathrm{T}}{\mu}}(\because$ Length of wire is made doubled)
For second harmonic of sonometer wire,
$2 f^{\prime}{ }_{1}=2 \times \frac{1}{2(2 \mathrm{~L})} \sqrt{\frac{\mathrm{T}}{\mu}}$
$=\frac{1}{2 \mathrm{~L}} \sqrt{\frac{\mathrm{T}}{\mu}}$
$=f$
Thus, given tuning fork will resonate with given sonometer.
$(i)$ initially in first harmonic (i.e. fundamental frequency).
$(ii)$ finally (when length of wire is doubled in second harmonic (i.e. first overtone)
