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

The Formula
The formula that ties these all together is:
f = \dfrac{1}{2L} \sqrt{\dfrac{T}{m}}
Where:
* f is the frequency
* L is the length of the string
* T is the tension
* m is the mass per unit length (or just mass if lengths are the same)
Solving the Problem
* Given:
* Two strings have the same length (L_1 = L_2)
* They have the same tension (T_1 = T_2)
* The ratio of their frequencies is 4:1 (f_1/f_2 = 4/1)
* We want to find: The ratio of their masses (m_1/m_2)
* Applying the Formula:
* For string 1: f_1 = \dfrac{1}{2L} \sqrt{\dfrac{T}{m_1}}
* For string 2: f_2 = \dfrac{1}{2L} \sqrt{\dfrac{T}{m_2}}
* Dividing the Equations:
* \dfrac{f_1}{f_2} = \dfrac{\dfrac{1}{2L} \sqrt{\dfrac{T}{m_1}}}{\dfrac{1}{2L} \sqrt{\dfrac{T}{m_2}}}
* Simplifying:
* \dfrac{f_1}{f_2} = \sqrt{\dfrac{m_2}{m_1}}
* Substituting the Frequency Ratio:
* \dfrac{4}{1} = \sqrt{\dfrac{m_2}{m_1}}
* Squaring both sides:
* \dfrac{16}{1} = \dfrac{m_2}{m_1}
* Taking the reciprocal to find m1/m2:
* \dfrac{m_1}{m_2} = \dfrac{1}{16}
Therefore, the masses of the strings are in the ratio of 1:16.
The correct answer is D) 1:16