A string is stretched between fixed points separated by $75.0\,\, cm.$ It is observed to have resonant frequencies of $420\,\, Hz$ and $315\,\, Hz$. There are no other resonant frequencies between these two. The lowest resonant frequency for this string is .... $Hz$
AIPMT 2015, Medium
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For a string fixed at both ends, the resonant frequencies are
$v_{n}=\frac{m v}{2 L}$ where $n=1,2,3, \ldots .$
The difference between two consecutive resonant frequencies is
$\Delta v_{n}=v_{n+1}-v_{n}=\frac{(n+1) v}{2 L}-\frac{n v}{2 L}=\frac{v}{2 L}$
which is also the lowest resonant frequency $(n=1)$
Thus the lowest resonant frequency for the given string
$=420 \mathrm{Hz}-315 \mathrm{Hz}=105 \mathrm{Hz}$
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