Two uniform strings A and B made of steel are made to vibrate under the same tension. If first overtone of A…

Two uniform strings A and B made of steel are made to vibrate under the same tension. If 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 length of string $B$ to that of $A$ is
  1. $1: 2$
  2. $4: 3$
  3. $2: 3$
  4. $3: 1$

Solution

$\mathrm{R}_2=2 \mathrm{R}_1, \mathrm{~T}_1=\mathrm{T}_2$
As the first overtone of A is equal to second overtone of B, we get $\begin{aligned} 2 n_A=3 n_B \\ \ldots(\because \text { first overtone }=\text { second } \end{aligned}$ harmonic)...(i) Also, $\mathrm{n}=\frac{1}{2 l} \sqrt{\frac{\mathrm{~T}}{\mathrm{~m}}}$. Where, $\mathrm{m}=$ mass per unit length $=\frac{\left(\pi \mathrm{R}^2 l\right) \rho}{l} . .(\mathrm{ii})$ $\begin{aligned} & \therefore \quad 2 \times \frac{1}{2 l_{\mathrm{A}}} \sqrt{\frac{\mathrm{~T}}{\mathrm{~m}_{\mathrm{A}}}}=3 \times \frac{1}{2 l_B} \sqrt{\frac{\mathrm{~T}}{\mathrm{~m}_{\mathrm{B}}}} ...(from(i))\\ & \therefore \quad \frac{l_{\mathrm{A}}}{l_{\mathrm{B}}}=\frac{2}{3} \sqrt{\frac{\mathrm{~m}_{\mathrm{B}}}{\mathrm{~m}_{\mathrm{A}}}}=\frac{2}{3} \sqrt{\frac{\pi \mathrm{r}_{\mathrm{B}}^2}{\pi \mathrm{r}_{\mathrm{A}}^2}} \\ & \therefore \quad \frac{l_{\mathrm{A}}}{l_{\mathrm{B}}}=\frac{2}{3} \sqrt{\frac{\mathrm{r}_{\mathrm{B}}^2}{\left(2 \mathrm{r}_{\mathrm{B}}\right)^2}}=\frac{2}{3} \sqrt{\frac{1}{4}}=\frac{1}{3} \\ & \therefore \quad \frac{l_{\mathrm{B}}}{l_{\mathrm{A}}}=\frac{3}{1} \end{aligned}$ ...(from (ii))

Asked in: MHT CET 2024 (16 May Shift 2)

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