Question

Two uniform wires of same material are vibrating under the same tension. If the \(1^{\text {st }}\) overtone of \(1^{\text {st }}\) wire is equal to the \(2^{\text {nd }}\) overtone of \(2^{\text {nd }}\) wire and radius of \(1^{\text {st }}\) wire is twice the radius of \(2^{\text {nd }}\) wire, the ratio of length of \(1^{\text {st }}\) wire to that \(2^{\text {nd }}\) wire is

• A. \(1:3\)
• B. \(3:1\)
• C. \(2:3\)
• D. \(3:5\)

Hint:

Frequency of a stretched wire depends on length and wave speed, and speed depends on the wire radius at fixed tension.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
Wave speed is \(v=\sqrt{T/\mu}\) with \(\mu=\rho\pi r^{2}\), so \(v\propto\dfrac{1}{r}\).
Frequency of the nth harmonic is \(f=\dfrac{nv}{2L}\).

Step 2: Speed ratio
\(\dfrac{v_1}{v_2}=\dfrac{r_2}{r_1}=\dfrac{1}{2}\)

Step 3: Equal frequencies
1st overtone of wire 1 (n = 2) equals 2nd overtone of wire 2 (n = 3):
\(\dfrac{2v_1}{2L_1}=\dfrac{3v_2}{2L_2}\)

Step 4: Calculation
\(\dfrac{L_1}{L_2}=\dfrac{2v_1}{3v_2}=\dfrac{2}{3}\times\dfrac{1}{2}=\dfrac{1}{3}\)

Final Answer:
The ratio of lengths is 1 : 3. \[ \boxed{\text{(A) }1:3} \]