TG EAMCET 3 May 2025 Shift 1 Paper

Step 1: Formula for Fundamental FrequencyThe basic formula for the frequency of a vibrating string is:$f=\;\frac{1}{2 l} \sqrt{\;\frac{T}{\mu}}$Here, $f$ is frequency, $l$ is the length of the string, $T$ is the tension, and $\mu$ is mass per unit length (which stays the same in this problem).Step 2: Frequency with Initial Length and TensionThe first frequency, with length $l$ and tension $T_1$ is:$f_1=\;\frac{1}{2 l} \sqrt{\;\frac{T_1}{\mu}}$Step 3: Frequency After Changing Length and TensionThe length of the string is shortened by $25 \%$, so the new length is $75 \%$ of $l$, or $\;\frac{3}{4} l$. The new tension is $T_2$ and the new frequency becomes $2 f_1$ (it increases by $100 \%$ ):$2 f_1=\;\frac{1}{2 \left( \;\frac{3}{4} l \right)} \sqrt{\;\frac{T_2}{\mu}}$Step 4: Set Up the EquationNow, make it easier by expressing $2 f_1$ in terms of the formula:$2 f_1=\;\frac{1}{\left( \frac{3}{2} \right) l} \sqrt{\;\frac{T_2}{\mu}}$Because splitting the denominator gives $2 \times \;\frac{1}{2 \left( \;\frac{3}{4} l \right)} = \;\frac{1}{\left( \frac{3}{2} \right) l}$.Step 5: Plug $f_1$ from Step 2 into the New EquationReplace $f_1$ in the new formula using the old frequency formula:$2 \left( \;\frac{1}{2 l} \sqrt{\;\frac{T_1}{\mu}} \right) = \;\frac{1}{\left( \frac{3}{2} \right) l} \sqrt{\;\frac{T_2}{\mu}}$Step 6: Simplify the EquationMultiply both sides to clear out the denominators: $\;\frac{1}{l} \sqrt{\;\frac{T_1}{\mu}} = \;\frac{2}{3 l} \sqrt{\;\frac{T_2}{\mu}}$Multiply both sides by $3 l$ :$3 \sqrt{\;\frac{T_1}{\mu}} = 2 \sqrt{\;\frac{T_2}{\mu}}$Step 7: Get Rid of Square Roots by Squaring Both SidesSquare each side to eliminate the square roots:$9 \;\frac{T_1}{\mu} = 4 \;\frac{T_2}{\mu}$The $\mu$ cancels out:$9 T_1=4 T_2$Step 8: Find $\;\frac{T_2}{T_1}$Divide both sides by $4 T_1$ to solve for $\;\frac{T_2}{T_1}$ :$\;\frac{T_2}{T_1} = \;\frac{9}{4}$