A string is stretched between two fixed points separated by $75\,cm$ . It is observed to have resonant frequencies of $420\,Hz$ and $315\,Hz$ . There are no other resonant frequencies between these two. Then, the lowest resonant frequency for this string is ..... $Hz$
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$\frac{\mathrm{n}}{2 l} \mathrm{v}=315 \quad$ and $\quad \frac{(\mathrm{n}+1) \mathrm{v}}{2 l}=420$
Solving $, \frac{\mathrm{v}}{2 l}=105 \mathrm{\,Hz}$
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