Question

The fundamental frequency of a sonometer wire increases by \(4\) Hz if the tension in the string is increased by \(21\%\) keeping the length of the wire constant. What will be the new fundamental frequency of the wire if its length is increased by \(25\%\) while keeping the original tension in the wire?

• A. Will remain the same
• B. \(32\) Hz
• C. \(34\) Hz
• D. \(35\) Hz

Hint:

For a sonometer wire: \[ f\propto \sqrt{T} \] and \[ f\propto \frac1L. \] A \(21\%\) increase in tension produces a \(10\%\) increase in frequency because \[ \sqrt{1.21}=1.1. \]

Answer 1

The Correct Option is B

Step 1: Recall how frequency depends on tension and length.
For a stretched wire $f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}$. So with length fixed, $f \propto \sqrt{T}$, and with tension fixed, $f \propto \frac{1}{L}$.
Step 2: Use the tension change to find the original frequency.
Increasing tension by $21\%$ makes it $1.21T$, so the new frequency is $f' = f\sqrt{1.21} = 1.1f$.
Step 3: Apply the given $4$ Hz rise.
The rise is $f' - f = 1.1f - f = 0.1f = 4$ Hz.
Step 4: Solve for the original frequency.
$0.1f = 4$ gives $f = 40$ Hz.
Step 5: Now change the length instead.
With the original tension restored, length is increased by $25\%$, so $L' = 1.25L = \frac{5}{4}L$. Since $f \propto \frac{1}{L}$, the new frequency is $f'' = f\cdot\frac{L}{L'} = 40\times\frac{4}{5}$.
Step 6: Finish the arithmetic.
\[ f'' = 40 \times 0.8 = 32\ \text{Hz} \]
\[ \boxed{f'' = 32\ \text{Hz}} \]