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A string in a musical instrument is $50 \mathrm{~cm}$ long and its fundamental frequency is $800 \mathrm{~Hz}$. If a frequency of $1000 \mathrm{~Hz}$ is to be produced, then required length of string is
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The correct answer is:
$40 \mathrm{~cm}$
Initial length of string $\left(l_1\right)=50 \mathrm{~cm}=0.5 \mathrm{~m}$ Initial fundamental frequency $\left(n_1\right)=800 \mathrm{~Hz}$ and final frequency $\left(n_2\right)=1000 \mathrm{~Hz}$.
Frequency of vibration of the string, $n=\frac{1}{2 l} \sqrt{\frac{T}{m}} \propto \frac{1}{l}$
$$
\therefore \quad \frac{n_1}{n_2}=\frac{l_2}{l_1} \quad \text { or } \quad l_2=\frac{n_1 \times l_1}{n_2}=\frac{800 \times 50}{1000}=40 \mathrm{~cm} \text {. }
$$
(where $l_2=$ final length of the string).
Frequency of vibration of the string, $n=\frac{1}{2 l} \sqrt{\frac{T}{m}} \propto \frac{1}{l}$
$$
\therefore \quad \frac{n_1}{n_2}=\frac{l_2}{l_1} \quad \text { or } \quad l_2=\frac{n_1 \times l_1}{n_2}=\frac{800 \times 50}{1000}=40 \mathrm{~cm} \text {. }
$$
(where $l_2=$ final length of the string).
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