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A string of length $100 \mathrm{~cm}$ has three resonant frequencies, $120 \mathrm{~Hz}, 200 \mathrm{~Hz}$ and $280 \mathrm{~Hz}$. If a node is formed at the end of the string, the speed of the transverse wave on this string is :
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The correct answer is:
$80 \mathrm{~m} / \mathrm{s}$
Given, length of string, $l=100 \mathrm{~cm}=1 \mathrm{~m}$
Three resonant frequencies are given as $f_1=120 \mathrm{~Hz}, f_2=200 \mathrm{~Hz}, f_3=280 \mathrm{~Hz}$
$\because$ Fundamental frequency $\left(f_0\right)$ is equal to the greatest common factor of $f_1, f_2$ and $f_3$.
$\therefore f_0=$ greatest common factor of $120 \mathrm{~Hz}, 200 \mathrm{~Hz}$ and $280 \mathrm{~Hz}$ i.e., $f_0=40 \mathrm{~Hz}$
Now, we know that, speed of transverse wave is given by $v=2 l f_0=2 \times 1 \times 40=80 \mathrm{~m} / \mathrm{s}$.
Three resonant frequencies are given as $f_1=120 \mathrm{~Hz}, f_2=200 \mathrm{~Hz}, f_3=280 \mathrm{~Hz}$
$\because$ Fundamental frequency $\left(f_0\right)$ is equal to the greatest common factor of $f_1, f_2$ and $f_3$.
$\therefore f_0=$ greatest common factor of $120 \mathrm{~Hz}, 200 \mathrm{~Hz}$ and $280 \mathrm{~Hz}$ i.e., $f_0=40 \mathrm{~Hz}$
Now, we know that, speed of transverse wave is given by $v=2 l f_0=2 \times 1 \times 40=80 \mathrm{~m} / \mathrm{s}$.
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