Q141 / 160 · AP EAMCET 2024 (22 May Shift 1) · Physics
The vibrations of four air columns are shown below. The ratio of frequencies is
- A1:2:3:4
- B1:3:2:4
- C1:4:3:2
- Correct answerD1:4:2:3
Solution
For fig(a), $L=\frac{\lambda_1}{4} \Rightarrow \lambda_1=4 \mathrm{~L}$
For fig (b), $L=\frac{\lambda_2}{4}+\frac{\lambda_2}{2}+\frac{\lambda_2}{4}=\lambda_2 \Rightarrow \lambda_2=\mathrm{L}$ For fig (c), $L=\frac{\lambda_3}{4}+\frac{\lambda_3}{4}+\frac{\lambda_3}{2} \Rightarrow \lambda_3=2 \mathrm{~L}$ For fig (d), $L=\frac{\lambda_4}{2}+\frac{\lambda_4}{4}=\frac{3 \lambda_4}{4} \Rightarrow \lambda_4=\frac{4 \mathrm{~L}}{3}$ For air column, $\mathrm{V} \propto \frac{1}{\lambda}$ $\begin{aligned} & \Rightarrow \mathrm{V}_1: \mathrm{V}_2: \mathrm{V}_3: \mathrm{V}_4=\frac{1}{\lambda_1}: \frac{1}{\lambda_2}: \frac{1}{\lambda_3}: \frac{1}{\lambda_4} \\ & =\frac{1}{4 \mathrm{~L}}: \frac{1}{\mathrm{~L}}: \frac{1}{2 \mathrm{~L}}: \frac{3}{4 \mathrm{~L}} \\ & =1: 4: 2: 3 \end{aligned}$
For fig (b), $L=\frac{\lambda_2}{4}+\frac{\lambda_2}{2}+\frac{\lambda_2}{4}=\lambda_2 \Rightarrow \lambda_2=\mathrm{L}$ For fig (c), $L=\frac{\lambda_3}{4}+\frac{\lambda_3}{4}+\frac{\lambda_3}{2} \Rightarrow \lambda_3=2 \mathrm{~L}$ For fig (d), $L=\frac{\lambda_4}{2}+\frac{\lambda_4}{4}=\frac{3 \lambda_4}{4} \Rightarrow \lambda_4=\frac{4 \mathrm{~L}}{3}$ For air column, $\mathrm{V} \propto \frac{1}{\lambda}$ $\begin{aligned} & \Rightarrow \mathrm{V}_1: \mathrm{V}_2: \mathrm{V}_3: \mathrm{V}_4=\frac{1}{\lambda_1}: \frac{1}{\lambda_2}: \frac{1}{\lambda_3}: \frac{1}{\lambda_4} \\ & =\frac{1}{4 \mathrm{~L}}: \frac{1}{\mathrm{~L}}: \frac{1}{2 \mathrm{~L}}: \frac{3}{4 \mathrm{~L}} \\ & =1: 4: 2: 3 \end{aligned}$
Hence, the correct answer is option: 'D'.