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

• A. $\frac{4}{9}$
• B. $\frac{2}{9}$
• C. $\frac{6}{4}$
• D. $\frac{9}{4}$

Hint:

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 $9:4$.

Answer 1

The Correct Option is D

Step 1: Frequency of a string.
The fundamental frequency is $n = \dfrac{1}{2L}\sqrt{\dfrac{T}{\mu}}$, where $T$ is tension and $\mu$ is mass per length.

Step 2: The strings are identical.
Same material and size means same $L$ and same $\mu$. So $n \propto \sqrt{T}$.

Step 3: Square both sides.
Then $n^{2} \propto T$, which gives $\dfrac{T_{A}}{T_{B}} = \left(\dfrac{n_{A}}{n_{B}}\right)^{2}$.

Step 4: Put in the frequencies.
$n_{A}=450$ Hz and $n_{B}=300$ Hz, so the ratio is $\dfrac{450}{300} = \dfrac{3}{2}$.

Step 5: Square the ratio.
\[ \frac{T_{A}}{T_{B}} = \left(\frac{3}{2}\right)^{2} = \frac{9}{4} \]
Step 6: Read the answer.
So $\dfrac{T_{A}}{T_{B}} = \dfrac{9}{4}$, which is option 4.
\[ \boxed{\frac{T_{A}}{T_{B}} = \frac{9}{4}} \]