MHT CET · Physics · Waves and Sound

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

A $\frac{1}{3}$
B $\frac{1}{4}$
C $\frac{1}{5}$
D $\frac{1}{6}$

Verified Solution

Answer & Solution

Correct Answer

(A) $\frac{1}{3}$

Step-by-step Solution

Detailed explanation

Fundamental frequency of the first wire

f = \frac{1}{2 L_{1}} \sqrt{\frac{T}{m}} = \frac{1}{2 L_{1}} \sqrt{\frac{T}{{π r}_{1}^{2} ρ}} = \frac{1}{2 L_{1} r_{1}} \sqrt{\frac{T}{π ρ}}

The first overtone

f_{1} = 2 f = \frac{2}{2 L_{1} r_{1}} \sqrt{\frac{T}{π ρ}} = \frac{1}{L_{1} r_{1}} \sqrt{\frac{T}{π ρ}}

The second overtone of the second wire

f_{2} = \frac{3}{2 L_{2} r_{2}} \sqrt{\frac{T}{π ρ}}

f_{1} = f_{2}

\frac{1}{L_{1} r_{1}} \sqrt{\frac{T}{π ρ}} = \frac{3}{2 L_{2} r_{2}} \sqrt{\frac{T}{π ρ}}

∴ 3 L_{1} r_{1} = 2 L_{2} r_{2}

\frac{L_{1}}{L_{2}} = \frac{2}{3} . \frac{r_{2}}{r_{1}}

= \frac{2}{3} . \frac{r_{2}}{2 r_{2}}

[∴ r_{1} = 2 r_{2}]

= \frac{1}{3}

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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