Consider the sound wave travelling in ideal gases of mathrmHe$\mathrm{He}$, mathrmCH_4$\mathrm{CH}_{4}$, and mathrmCO_2$\mathrm{CO}_{2}$. All the gases have the same ratio fracP ho$\frac{P}{ ho}$, where P$P$ is the pressure and ho$ ho$ is the density. The ratio of the speed of sound through the gases v_mathrmHe : v_mathrmCH_4 : v_mathrmCO_2$v_{\mathrm{He}} : v_{\mathrm{CH}_{4}} : v_{\mathrm{CO}_{2}}$ is given by

### Related Formula f_textclosed, fundamental = fracv4L_c$$f_{\text{closed, fundamental}} = \frac{v}{4L_c}$$ f_textopen, 1st overtone = frac2v2L_o$$f_{\text{open, 1st overtone}} = \frac{2v}{2L_o}$$ ### Core Logic

For a closed organ pipe, the fundamental frequency (1st harmonic) is: f_1 = fracvlambda = fracv4L_1$$f_1 = \frac{v}{\lambda} = \frac{v}{4L_1}$$ where L_1$L_1$ is the length of the closed pipe. For an open organ pipe, the first overtone (2nd harmonic) is: f_2 = frac2v2L_2 = fracvL_2$$f_2 = \frac{2v}{2L_2} = \frac{v}{L_2}$$ where L_2$L_2$ is the length of the open pipe (L_2 = 60mathrm\,cm$L_2 = 60\mathrm{\,cm}$). ### Step 2: Equating Frequencies Given f_1 = f_2$f_1 = f_2$: fracv4L_1 = fracvL_2$$\frac{v}{4L_1} = \frac{v}{L_2}$$ L_2 = 4L_1$L_2 = 4L_1$ 60 = 4 times L_1$$60 = 4 \times L_1$$ L_1 = 15mathrm\,cm$$L_1 = 15\mathrm{\,cm}$$ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Waves