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A string is stretched between two rigid supports separated by $75 \mathrm{~cm}$. There are no resonant frequencies between $420 \mathrm{~Hz}$ and $315 \mathrm{~Hz}$. The lowest resonant frequency for the string is
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Verified Answer
The correct answer is:
$105 \mathrm{~Hz}$
As there is no resonant frequency between $315 \mathrm{~Hz}$ and $420 \mathrm{~Hz}$, let $315 \mathrm{~Hz}$ be $\mathrm{n}^{\text {th }}$ overtone and $420 \mathrm{~Hz}$ be $(\mathrm{n}+1)^{\text {th }}$ overtone.
Now, $v=\frac{\text { nv }}{2 l}$
$$
\therefore \quad 315=\frac{\mathrm{nv}}{2 l} \text { and } 420=\frac{(\mathrm{n}+1) \mathrm{v}}{2 l}
$$
Taking the ratio,
$$
\begin{array}{ll}
& \frac{315}{420}=\frac{n}{n+1} \\
\therefore \quad & 315 n+315=420 n \\
\therefore \quad & n=3
\end{array}
$$
The resonant frequency is $v_0=\frac{\mathrm{v}}{2 l}$
Therefore, from equation (i) we get, $v_0=\frac{v}{n}=\frac{315}{3}=105 \mathrm{~Hz}$
Now, $v=\frac{\text { nv }}{2 l}$
$$
\therefore \quad 315=\frac{\mathrm{nv}}{2 l} \text { and } 420=\frac{(\mathrm{n}+1) \mathrm{v}}{2 l}
$$
Taking the ratio,
$$
\begin{array}{ll}
& \frac{315}{420}=\frac{n}{n+1} \\
\therefore \quad & 315 n+315=420 n \\
\therefore \quad & n=3
\end{array}
$$
The resonant frequency is $v_0=\frac{\mathrm{v}}{2 l}$
Therefore, from equation (i) we get, $v_0=\frac{v}{n}=\frac{315}{3}=105 \mathrm{~Hz}$
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