Physics - Wave on String Question with Solution | TestHub

Frequency is given by $\frac{v}{\lambda }$, where $v$ is the speed of the sound and $\lambda$ the length of the pipe. For the second overtone of an open pipe: $\begin{array}{rl}& \frac{3}{2}\lambda =L_O\\ & \therefore f_O=\frac{3v}{2L_O}\end{array}$ For the first overtone of the closed pipe: $\begin{array}{rl}& \frac{3}{4}\lambda =L_C\\ & \therefore f_C=\frac{3v}{4L_C}\end{array}$ If they have the same frequency: $\frac{3v}{2L_O}=\frac{3v}{4L_C}$ $\Rightarrow L_C=\frac{L_O}{2}=\frac{1.5}{2}\mathrm{m}$