Two strings of the same length and under the same tension...
Two strings of the same length and under the same tension gives notes of frequencies in the ratio 4 : 1. The masses of the strings are in the corresponding ratio of
2 : 1
1 : 2
1 : 4
1 : 16
Correct answer is D
F = \(\frac{1}{2L}\)\(\sqrt\frac{T}{m}\)
\(\frac{F_1}{F_2}\) = \(\sqrt\frac{m_2}{m_1}\)
\(\frac{1}{4}\) = \(\sqrt\frac{m_2}{m_1}\)
\(\frac{m_2}{m_1}\) = \(\frac{1}{16}\)