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Two strings of the same material and the same area of cross-section are used in sonometer experiment. One is loaded with 12kg and the other with 3kg. The fundamental frequency of the first string is equal to the first overtone of the second string. If the length of the second string is 100cm, then the length of the first string is …..cm

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Detailed Solution

we know frequency in stretched string is given as

$f_{n} = \frac{n}{2 l} \sqrt{\frac{T}{ρA}}$

From the given condition :

$f_{0} = f'_{1}$

$\Rightarrow \frac{1}{2 l_{1}} \sqrt{\frac{T_{1}}{ρA}} = \frac{2}{2 l_{2}} \sqrt{\frac{T_{2}}{ρA}}$

$\Rightarrow l_{2} = 2 l_{1} \sqrt{\frac{T_{2}}{T_{1}}} = 2 l_{1} \sqrt{\frac{3 \times 10}{12 \times 10}} = l_{1}$

Therefore length of the first string is 100cm

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