Two identical strings of length ell and 2ell | vedclass

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(C) The fundamental frequency of a stretched string is given by $f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$,where $L$ is the length,$T$ is the tension,and

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$1$:$9$

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(C) The fundamental frequency of a stretched string is given by $f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$,where $L$ is the length,$T$ is the tension,and $\mu$ is the mass per unit length.Since the strings are identical,$\mu$ is the same for both.For the first string: $N = \frac{1}{2\ell} \sqrt{\frac{T_1}{\mu}}$For the second string: $1.5N = \frac{1}{2(2\ell)} \sqrt{\frac{T_2}{\mu}} = \frac{1}{4\ell} \sqrt{\frac{T_2}{\mu}}$Dividing the two equations: $\frac{N}{1.5N} = \frac{\frac{1}{2\ell} \sqrt{\frac{T_1}{\mu}}}{\frac{1}{4\ell} \sqrt{\frac{T_2}{\mu}}}$$\frac{1}{1.5} = \frac{4\ell}{2\ell} \sqrt{\frac{T_1}{T_2}}$$\frac{1}{1.5} = 2 \sqrt{\frac{T_1}{T_2}}$$\sqrt{\frac{T_1}{T_2}} = \frac{1}{1.5 	imes 2} = \frac{1}{3}$Squaring both sides: $\frac{T_1}{T_2} = \frac{1}{9}$.

Two identical strings of length $\ell$ and $2\ell$ vibrate with fundamental frequencies $N$ Hz and $1.5N$ Hz,respectively. The ratio of tensions for the smaller length to the larger length is

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