Question 389: Waves

Question

Two uniform wires of same material are vibrating under the same tension. If the first overtone of first wire is equal to the $$2^{	ext {nd }}$$ overtone of $$2^{	ext {nd }}$$ wire and radius of the first wire is twice the radius of the $$2^{	ext {nd }}$$ wire then the ratio of length of first wire to $$2^{	ext {nd }}$$ wire is

Accepted Answer

Fundamental frequency of the first wire is
$$\mathrm{n}=\frac{1}{2 l_1} \sqrt{\frac{\mathrm{T}}{\mathrm{m}}}=\frac{1}{2 l_1} \sqrt{\frac{\mathrm{T}}{\pi \mathrm{r}_1^2 ho}}=\frac{1}{2 l_1 \mathrm{r}_1} \sqrt{\frac{\mathrm{T}}{\pi ho}}$$
The first overtone $$n_1=2 n=\frac{1}{l_1 r_1} \sqrt{\frac{T}{\pi ho}}$$
Similarly, the second overtone of the second wire will be,
$$\mathrm{n}_2=\frac{3}{2 l_2 \mathrm{r}_2} \sqrt{\frac{\mathrm{T}}{\pi ho}}$$
Given that $$\mathrm{n}_1=\mathrm{n}_2$$
$$\begin{aligned}
& 	herefore \quad \frac{1}{l_1 \mathrm{r}_1} \sqrt{\frac{\mathrm{T}}{\pi ho}}=\frac{3}{2 l_2 \mathrm{r}_2} \sqrt{\frac{\mathrm{T}}{\pi ho}} \
& 	herefore \quad 3 l_1 \mathrm{r}_1=2 l_2 \mathrm{r}_2 \
& \frac{l_1}{l_2}=\frac{2 r_2}{3 r_1} \
& =\frac{2 r_2}{3\left(2 r_2ight)} \quad\ldots\left(\because r_1=2 r_2ight) \
& =\frac{1}{3} \
\end{aligned}$$

Answer

$$1: 3$$

Answer

$$3: 1$$

Answer

$$1: 9$$

Answer

$$9: 1$$

Waves

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