Question

Two wires of same material are vibrating under the same tension. If the first overtone of first wire is equal to the second overtone of second wire and radius of first wire is twice the radius of the second wire, then the ratio of length of first wire to second wire is

• A. \( 1 : 3 \)
• B. \( 1 : 2 \)
• C. \( 2 : 1 \)
• D. \( 3 : 1 \)

Hint:

For strings of same material, linear density varies as square of radius.

Answer 1

The Correct Option is A

Step 1: Write frequency formula for stretched string.
\[ f_n = \frac{n}{2L}\sqrt{\frac{T}{\mu}} \]
Step 2: Identify overtones.
First overtone \( = 2f_1 \), second overtone \( = 3f_1 \).
Step 3: Use given condition.
\[ \frac{2}{2L_1}\sqrt{\frac{T}{\mu_1}} = \frac{3}{2L_2}\sqrt{\frac{T}{\mu_2}} \]
Step 4: Relate linear density.
\[ \mu \propto r^2 \Rightarrow \mu_1 = 4\mu_2 \]
Step 5: Simplify.
\[ \frac{1}{L_1}\cdot\frac{1}{2} = \frac{3}{L_2} \Rightarrow \frac{L_1}{L_2} = \frac{1}{3} \]
Step 6: Conclusion.
The ratio of lengths is \( 1 : 3 \).