Question:
Two identical strings A and B of same material are having tensions $T_{A}$ and $T_{B}$ respectively. If their fundamental frequencies are 450 Hz and 300 Hz, then $T_{A}/T_{B}$ is
Two identical strings A and B of same material are having tensions $T_{A}$ and $T_{B}$ respectively. If their fundamental frequencies are 450 Hz and 300 Hz, then $T_{A}/T_{B}$ is
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Frequency is directly proportional to the square root of tension ($\sqrt{T}$). If the frequency ratio simplifies to $3:2$, then squaring it instantly gives the tension ratio as $9:4$.
Updated On: Jun 3, 2026
- $\frac{4}{9}$
- $\frac{2}{9}$
- $\frac{6}{4}$
- $\frac{9}{4}$
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The Correct Option is D
Solution and Explanation
Step 1: Concept
The fundamental frequency ($n$) of a stretched string fixed at both ends is given by the formula $n = \frac{1}{2L}\sqrt{\frac{T}{\mu}}$, where $L$ is the length of the string, $T$ is the tension, and $\mu$ is the linear mass density.
Step 2: Meaning
Since both strings are identical and composed of the exact same material, their lengths ($L$) and linear mass densities ($\mu$) are identical. Therefore, the fundamental frequency is directly proportional to the square root of the tension: $n \propto \sqrt{T}$.
Step 3: Analysis
Squaring both sides gives $n^2 \propto T$, which establishes the ratio equation: $\frac{T_{A}}{T_{B}} = \left(\frac{n_{A}}{n_{B}}\right)^2$. Given that $n_{A} = 450 \text{ Hz}$ and $n_{B} = 300 \text{ Hz}$, we can substitute these values into the ratio: $\frac{T_{A}}{T_{B}} = \left(\frac{450}{300}\right)^2 = \left(\frac{3}{2}\right)^2 = \frac{9}{4}$.
Step 4: Conclusion
Hence, the required ratio of the tensions in strings A and B is $\frac{9}{4}$.
Final Answer: (D)
The fundamental frequency ($n$) of a stretched string fixed at both ends is given by the formula $n = \frac{1}{2L}\sqrt{\frac{T}{\mu}}$, where $L$ is the length of the string, $T$ is the tension, and $\mu$ is the linear mass density.
Step 2: Meaning
Since both strings are identical and composed of the exact same material, their lengths ($L$) and linear mass densities ($\mu$) are identical. Therefore, the fundamental frequency is directly proportional to the square root of the tension: $n \propto \sqrt{T}$.
Step 3: Analysis
Squaring both sides gives $n^2 \propto T$, which establishes the ratio equation: $\frac{T_{A}}{T_{B}} = \left(\frac{n_{A}}{n_{B}}\right)^2$. Given that $n_{A} = 450 \text{ Hz}$ and $n_{B} = 300 \text{ Hz}$, we can substitute these values into the ratio: $\frac{T_{A}}{T_{B}} = \left(\frac{450}{300}\right)^2 = \left(\frac{3}{2}\right)^2 = \frac{9}{4}$.
Step 4: Conclusion
Hence, the required ratio of the tensions in strings A and B is $\frac{9}{4}$.
Final Answer: (D)
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