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Superposition of Waves — Physics STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 SciencePhysicsSuperposition of Waves4 Marks
Question
A stretched sonometer wire emits a fundamental note of frequency $256 Hz$. Keeping the stretching force constant and reducing the length of the wire by $10 cm$, the frequency becomes $320 Hz$. Calculate the original length of the wire.
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Answer

Data: $n _1=256 Hz , T$ and $m$ constant, $L _2= L _1-10 cm , n _2=320 Hz$
$n=\frac{1}{2 L} \sqrt{\frac{T}{m}} \text { MaharashtraBoardSolutions.Guru }$
$ \therefore 256=\frac{1}{2 L_1} \sqrt{\frac{T}{m}} \text { and } 320=\frac{1}{2\left(L_1-10\right)} \sqrt{\frac{T}{m}}$
$\therefore \frac{256}{320}=\frac{L_1-10}{L_1} \quad \therefore \frac{4}{5}=\frac{L_1-10}{L_1}$
$\therefore 5 L_1-50=4 L_1$
$\therefore L_1=50 cm =0.5 m $
Alternative method:
Since $T$ and $m$ are constant, $nL =$ constant.
$ \therefore n _1 L _1= n _2 L _2 \therefore \frac{L_1}{L_2}=\frac{n_2}{n_1}$
$\therefore \frac{L_1}{L_1-10}=\frac{320}{256}=\frac{20}{16}=\frac{5}{4}$
$\therefore 4 L _1=5 L _1-50$
$\therefore 5 L _1-4 L _1=50$
$\therefore L _1=50 cm =0.5 m $

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