Answer:
The fundamental frequency of the stretched string is:
[tex]f=[/tex] [tex]\frac{1}{2} \sqrt{\frac{T}{L} }[/tex] [ T = Tension and μ = mass per unit length]
Here,
μ = [tex]\frac{m}{L} = \frac{Vp}{L} = Ap[/tex]
[tex]f= \frac{1}{2} \sqrt{\frac{T}{Ap} }[/tex]
If T is halved and A is doubled,
[tex]f= \frac{1}{2} \sqrt{\frac{T'}{A'p} } = \sqrt{\frac{1}{2* 2* A* p} } = \frac{1}{2} (\frac{1}{2} \sqrt{\frac{T}{Ap} } = \frac{1}{2} f[/tex]
Thus, the frequency is reduced to half if its tension is halved and the area of cross-section of the string is doubled.
When the tension is halved and the area of cross section is doubled, the frequency increases by a factor of 1.414.
The frequency of transverse vibration of a stretched string is calculated as follows;
[tex]f = \frac{1}{2l} \sqrt{\frac{T}{\mu} }[/tex]
where;
[tex]f = \frac{1}{2l} \sqrt{\frac{T}{\mu} }\\\\f = \frac{1}{2l} \sqrt{\frac{Tl}{m} } \\\\f = \frac{l^2}{2l} \sqrt{\frac{T}{m} } \\\\f = \frac{A}{2l} \sqrt{\frac{T}{m}}[/tex]
when the tension is halved and the area of cross section is doubled, the frequency is calculated as;
[tex]\frac{f_1}{A_1 \sqrt{T_1} } = \frac{f_2}{A_2\sqrt{T_2} } \\\\f_2 = \frac{f_1 A_2\sqrt{T_2}}{A_1 \sqrt{T_1} }\\\\f_2 = \frac{f_1 \times 2A_1\sqrt{0.5T_1} }{\sqrt{T_1} }\\\\f_2 = \frac{f_1 \times 2A_1 \times 0.7071\sqrt{T_1} }{\sqrt{T_1} }\\\\f_2 = 1.414 f_1[/tex]
Thus, when the tension is halved and the area of cross section is doubled, the frequency increases by a factor of 1.414.
Learn more about tension in a string here: https://brainly.com/question/25743940
