Physics - Sound Wave Question with Solution | TestHub

Let the length of closed organ pipe 1 be. Then, $$v_n=(2n-1)\frac{v}{4l}$$ Its third harmonics (put $n=2$ ) in above question. $$v_3=\frac{3}{4}\frac{v}{1}$$ Now, the length of open organ pipe is 21 , then $$v_n=n\frac{v}{2(2l)}=\frac{nv}{4l}$$ Its fundamental frequency, $$v_0=\frac{v}{2l}=100$$ $$v_3=\frac{3v}{4l}=3\times 100=300\mathrm{Hz}$$