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Question

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

Vedclass · Accepted Answer

$\frac{\Delta T}{T}=\left(\frac{f_0+N}{f_0}\right)^2-1$

Vedclass · Answer

(A) The fundamental frequency of a stretched wire is given by $f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$, where $T$ is the tension and $\mu$ is the linear mass density.Since the wires are identical, $L$ and $\mu$ are constant, so $f \propto \sqrt{T}$.Initially, for both wires, the frequency is $f_0 \propto \sqrt{T}$. Thus, $f_0^2 \propto T$ ... $(i)$When the tension of one wire is increased by $\Delta T$, its new frequency becomes $f' = f_0 + N$.Therefore, $(f_0 + N)^2 \propto (T + \Delta T)$ ... (ii)Dividing equation (ii) by equation $(i)$, we get:$\frac{(f_0 + N)^2}{f_0^2} = \frac{T + \Delta T}{T}$$\frac{(f_0 + N)^2}{f_0^2} = 1 + \frac{\Delta T}{T}$Rearranging for $\frac{\Delta T}{T}$:$\frac{\Delta T}{T} = \frac{(f_0 + N)^2}{f_0^2} - 1 = \left(\frac{f_0 + N}{f_0}\right)^2 - 1$.

Two identical wires have a fundamental frequency $f_0$ when kept under the same tension $T$. If the tension of one wire is increased by $\Delta T$, then $N$ beats occur when both wires oscillate simultaneously. Find the expression for $\frac{\Delta T}{T}$.

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