A sonometer's fundamental note is 50Hz, what is the new frequency when the tension is four times the original?
For a sonometer, F \(\propto\) \(\sqrt{T}\)
F = k \(\sqrt{T}\)
F\(_1\) = 50Hz, F\(_2\) = ?
T\(_1\) = T, T\(_2\) = 4T
\(\frac{F_1}{\sqrt{T_1}} = \frac{F_2}{\sqrt{T_2}}\)
\(\frac{50}{\sqrt{T}} = \frac{F_2}{\sqrt{4T}}\)
F\(_2\) = \(\frac{50 \times 2 \times\sqrt{T}}{\sqrt{T}}\)
F\(_2\) = 100Hz
There is an explanation video available below.
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