Complete Question
The complete question is shown on the first uploaded image
Answer:
The answer is
[tex]T_2 = 1.008[/tex] % higher than [tex]T_1[/tex]
[tex]T_2 = 0.99[/tex] % lower than [tex]T_1[/tex]
Explanation:
From the question we are told that
The first string has a frequency of [tex]f_1 = 230 Hz[/tex]
The period of the beat is [tex]t_{beat} = 0.99s[/tex]
Generally the frequency of the beat is
[tex]f_{beat} = \frac{1}{t_{beat}}[/tex]
Substituting values
[tex]f_{beat} = \frac{1}{0.99}[/tex]
[tex]= 1.01 Hz[/tex]
From the question
[tex]f_2 - f_1 = f_{beat}[/tex] for [tex]f_2[/tex] having a higher tension
So
[tex]f_2 - 230 = 1.01[/tex]
[tex]f_2 = 231.01Hz[/tex]
From the question
[tex]\frac{f_2}{f_1} = \sqrt{\frac{T_2}{T_1} }[/tex]
[tex]\frac{T_2}{T_1} = \frac{f_2^2}{f_1^2}[/tex]
Substituting values
[tex]\frac{T_2}{T_1} = \frac{(231.01)^2}{(230)^2}[/tex]
[tex]T_2 = 1.008[/tex] % higher than [tex]T_1[/tex]
For [tex]f_2[/tex] having a lower tension
[tex]f_1 - f_2 = f_{beat}[/tex]
So
[tex]230 - f_2 = 1.01[/tex]
[tex]f_2 = 230 -1.01[/tex]
[tex]= 228.99[/tex]
From the question
[tex]\frac{f_2}{f_1} = \sqrt{\frac{T_2}{T_1} }[/tex]
[tex]\frac{T_2}{T_1} = \frac{f_2^2}{f_1^2}[/tex]
Substituting values
[tex]\frac{T_2}{T_1} = \frac{(228.99)^2}{(230)^2}[/tex]
[tex]T_2 = 0.99[/tex] % lower than [tex]T_1[/tex]
