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Question

Two identical wires are stretched by the same tension of 100 N, and each emits a note of frequency 200 cycles/sec. The tension in one wire is increased by 1 N. Calculate the number of beats heard per second when the wires are plucked.

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Solution

Fundamental frequency $f = \frac{1}{2 L} \sqrt{\frac{T}{μ}}$
$\begin{array}{l} \frac{d f}{d T} = \frac{1}{2 L} [\frac{1}{2} {(\frac{T}{μ})}^{- 1 / 2} \frac{1}{μ}] \\ \Rightarrow \frac{d f}{d T} = \frac{1}{4 L T} {(\frac{T}{μ})}^{+ 1 / 2} = \frac{f}{2 T} \end{array}$
$\Rightarrow △ f = \frac{f}{2} \cdot \frac{ΔT}{T} = \frac{200}{2} \cdot \frac{1}{100} = 1 c y c l e / s e c .$

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Beats

Remember concepts with our Masterclasses.

Two identical wires are stretched by the same tension of 100 N, and each emits a note of frequency 200 cycles/sec. The tension in one wire is increased by 1 N. Calculate the number of beats heard per second when the wires are plucked.

Fundamental frequency f=12LTμ dfdT=12L12Tμ−1/21μ ​⇒dfdT=14LTTμ+1/2=f2T ⇒△f=f2⋅ΔTT=2002⋅1100= 1 cycle/sec.

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