When tension T is applied to a sonometer wire | vedclass

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(A) The fundamental frequency $n$ of a sonometer wire is given by the formula:$n = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$  ---$(1)$where $T$ is the tensio

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(A) The fundamental frequency $n$ of a sonometer wire is given by the formula:$n = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$  ---$(1)$where $T$ is the tension and $\mu$ is the linear mass density.When the tension is increased by $8 \,N$, the new tension becomes $(T + 8) \,N$ and the new frequency becomes $3n$. Thus:$3n = \frac{1}{2l} \sqrt{\frac{T + 8}{\mu}}$  ---$(2)$Dividing equation $(2)$ by equation $(1)$, we get:$\frac{3n}{n} = \frac{\frac{1}{2l} \sqrt{\frac{T + 8}{\mu}}}{\frac{1}{2l} \sqrt{\frac{T}{\mu}}}$$3 = \sqrt{\frac{T + 8}{T}}$Squaring both sides:$9 = \frac{T + 8}{T}$$9T = T + 8$$8T = 8$$T = 1 \,N$Therefore, the initial tension applied to the wire was $1 \,N$.

When tension $T$ is applied to a sonometer wire of length $l$, it vibrates with the fundamental frequency $n$. Keeping the setup same, when the tension is increased by $8 \,N$, the fundamental frequency becomes three times the earlier. The initial tension applied to the wire was: (in $\,N$)

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