When the tension in a sonometer wire is doubled, the rati...
When the tension in a sonometer wire is doubled, the ratio of the new frequency to the initial frequency is?
\(\frac{1}{\sqrt2}\)
\(\frac{1}{2}\)
\(\sqrt2\)
2
Correct answer is A
\(\frac{F_1}{F_2}\) = \(\sqrt{\frac{T_1}{T_2}}\)
\(T_2 = 2T_1\)
\(\frac{F_1}{F_2}\) = \(\sqrt{\frac{T_1}{2T_1}}\)
\(\frac{F_1}{F_2}\) = \(\sqrt2\)
F\(_2\) : F\(_1\) = 1 : \(\sqrt{2}\)
\(\frac{F_2}{F_1}\) = \(\frac{1}{\sqrt2}\)
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