A string is stretched between fixed points separated by 75.0 cm. … - Vidyadip

Question

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. Then, the lowest resonant frequency for this string is .... $Hz$

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Answer

Given $\frac{\mathrm{nv}}{2 \ell}=315$ and $(\mathrm{n}+1) \frac{\mathrm{v}}{2 \ell}=420$

$\Rightarrow \frac{n+1}{n}=\frac{420}{315} \Rightarrow n=3$

Hence $3 \times \frac{v}{2 \ell}=315 \Rightarrow \frac{v}{2 \ell}=105 \mathrm{Hz}$

Lowest resonant frequency is when $n=1$

Therefore lowest resonant frequency $=105 \mathrm{Hz}$

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