Question

Two uniform strings A and B made of steel are vibrating under the same tension. If the first overtone of A is equal to the second overtone of B and if the radius of A is twice that of B, the ratio of the lengths of the strings is

• A. \(2:1\)
• B. \(3:4\)
• C. \(3:2\)
• D. \(1:3\)

Hint:

Mass per unit length depends on the square of the radius, so the wave speed differs; match the overtone frequencies.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
For a string fixed at both ends, \(f_{n}=\dfrac{nv}{2L}\) with \(v=\sqrt{\dfrac{T}{\mu}}\) and \(\mu=\rho\pi r^{2}\).

Step 2: Wave Speeds
Same tension, and radius of A is twice B, so \(\mu_{A}=4\mu_{B}\) and \(v_{A}=\dfrac{v_{B}}{2}\).

Step 3: Match Overtones
First overtone of A is the second harmonic: \(\dfrac{2v_{A}}{2L_{A}}\). Second overtone of B is the third harmonic: \(\dfrac{3v_{B}}{2L_{B}}\).
Equating: \(\dfrac{v_{B}/2}{L_{A}}=\dfrac{3v_{B}}{2L_{B}}\Rightarrow\dfrac{L_{A}}{L_{B}}=\dfrac{1}{3}\)

Step 4: Compare the options
The ratio is \(1:3\), option (D).

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