A string is stretched between fixed points separated by $75.0 \mathrm{~cm}$. It is observed to have resonant frequencies of $420 \mathrm{~Hz}$ and $315 \mathrm{~Hz}$. There are no other resonant frequency for this string is: (1) $105 \mathrm{~Hz}$, (2) $155 \mathrm{~Hz}$ (3) $205 \mathrm{~Hz}$ (4) $105 \mathrm{~Hz}$

Home » A string is stretched between fixed points separated by . It is observed to have resonant frequencies of and . There are no other resonant frequency for this string is: (1) , (2) (3) (4)

A string is stretched between fixed points separated by $75.0 \mathrm{~cm}$ . It is observed to have resonant frequencies of $420 \mathrm{~Hz}$ and $315 \mathrm{~Hz}$ . There are no other resonant frequency for this string is: (1) $105 \mathrm{~Hz}$ , (2) $155 \mathrm{~Hz}$ (3) $205 \mathrm{~Hz}$ (4) $105 \mathrm{~Hz}$

Physics | Physics

A string is stretched between fixed points separated by $75.0 \mathrm{~cm}$ . It is observed to have resonant frequencies of $420 \mathrm{~Hz}$ and $315 \mathrm{~Hz}$ . There are no other resonant frequency for this string is: (1) $105 \mathrm{~Hz}$ , (2) $155 \mathrm{~Hz}$ (3) $205 \mathrm{~Hz}$ (4) $105 \mathrm{~Hz}$

The correct Solution is (1)
Two consecutive resonant frequencies for a string fixed at both ends will be
$\begin{array}{l} \frac{n v}{2 \ell} \text { and } \frac{(n+1) v}{2 \ell} \\ \Rightarrow \frac{(n+1) v}{2 \ell}-\frac{n v}{2 \ell}=420-315 \\ \frac{v}{2 \ell}=105 \mathrm{~Hz} \end{array}$
that is the minimum resonant frequency.

i

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