Question
In an experiment, it was found that a tuning fork and a sonometer wire gave 5 beats per second, both when the length of the wire was 1m, and 1.05m. Calculate the frequency of the fork.
✓
Answer
We know $\text{v}=\frac{1}{2\text{l}}\sqrt{\frac{\text{T}}{\mu}}$
I case: $\text{v}_1=\frac{1}{2\times1}\sqrt{\frac{\text{T}}{\mu}}\ ... (\text{i})$
II case: $\text{v}_2=\frac{1}{2\times1.05}\sqrt{\frac{\text{T}}{\mu}}\ ...\text{(ii)}$
Let v be the frequecy of the tuning fork, therefore,
$\text{v}_1-\text{v}=5$ and $\text{v}-\text{v}_2=5$
$\text{v}_1=\text{v}+5$
and $\text{v}_2=\text{v}-5$
$\frac{\text{v}_1}{\text{v}_2}=\frac{5+\text{v}}{\text{v}-5}$
From (i) and (ii), we have
$\frac{\text{v}+5}{\text{v}-5}=\frac{1}{2}\times2\times1.05=1.05$
$\frac{\text{v}+5}{\text{v}-5}=\frac{105}{100}=\frac{21}{20}$
$\Rightarrow\text{v}=205\text{Hz}$
I case: $\text{v}_1=\frac{1}{2\times1}\sqrt{\frac{\text{T}}{\mu}}\ ... (\text{i})$
II case: $\text{v}_2=\frac{1}{2\times1.05}\sqrt{\frac{\text{T}}{\mu}}\ ...\text{(ii)}$
Let v be the frequecy of the tuning fork, therefore,
$\text{v}_1-\text{v}=5$ and $\text{v}-\text{v}_2=5$
$\text{v}_1=\text{v}+5$
and $\text{v}_2=\text{v}-5$
$\frac{\text{v}_1}{\text{v}_2}=\frac{5+\text{v}}{\text{v}-5}$
From (i) and (ii), we have
$\frac{\text{v}+5}{\text{v}-5}=\frac{1}{2}\times2\times1.05=1.05$
$\frac{\text{v}+5}{\text{v}-5}=\frac{105}{100}=\frac{21}{20}$
$\Rightarrow\text{v}=205\text{Hz}$
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