1. **Problem (96):** A string fixed at one end acts like a pipe closed at one end with length $L=17$ cm. Given the speed of sound in air $v=340$ m/s, find the fundamental frequency. 2. **Step 1:** Convert length to meters: $L=0.17$ m. 3. **Step 2:** For a pipe closed at one end, the fundamental wavelength is $\lambda=4L$. 4. **Step 3:** Calculate wavelength: $\lambda=4 \times 0.17=0.68$ m. 5. **Step 4:** Frequency is $f=\frac{v}{\lambda}=\frac{340}{0.68}=500$ Hz. 6. **Problem (97):** A pipe of length $L$ contains air vibrating in its fundamental mode. Given speed of sound $V$, find the frequency of vibration. 7. **Step 1:** For a pipe closed at one end, fundamental wavelength $\lambda=4L$. 8. **Step 2:** Frequency $f=\frac{V}{\lambda}=\frac{V}{4L}$. 9. **Problem (98):** In an experiment with a pipe open at both ends, the first resonance occurs at length $L_1=10$ cm and the second resonance at $L_2=32$ cm. Find the fundamental wavelength. 10. **Step 1:** For an open pipe, resonances occur at $L_n= n \frac{\lambda}{2}$ where $n=1,2,3...$. 11. **Step 2:** Using $L_1=\frac{\lambda}{2}=0.10$ m and $L_2=\lambda=0.32$ m. 12. **Step 3:** These lengths do not fit the exact multiples, so find fundamental wavelength $\lambda$ by difference: $$L_2 - L_1 = \frac{3\lambda}{2} - \frac{\lambda}{2} = \lambda = 0.32 - 0.10 = 0.22 \text{ m}$$ 13. **Step 4:** So, fundamental wavelength $\lambda=0.22$ m. 14. **Problem (99):** Two pipes, one closed with length $L_1$ and one open with length $L_2$, have the same fundamental frequency. Find the ratio $\frac{L_1}{L_2}$. 15. **Step 1:** Frequency for closed pipe: $f=\frac{v}{4L_1}$. 16. **Step 2:** Frequency for open pipe: $f=\frac{v}{2L_2}$. 17. **Step 3:** Set equal: $\frac{v}{4L_1} = \frac{v}{2L_2}$. 18. **Step 4:** Simplify: $\frac{1}{4L_1} = \frac{1}{2L_2} \Rightarrow 2L_2 = 4L_1 \Rightarrow \frac{L_1}{L_2} = \frac{1}{2}$. **Final answers:** (96) $f=500$ Hz (97) $f=\frac{V}{4L}$ (98) $\lambda=0.22$ m (99) $\frac{L_1}{L_2}=\frac{1}{2}$