Question 138: Waves

Question

The fundamental frequency of a closed organ pipe is equal to the first overtone frequency of an open organ pipe. If length of the open pipe is $$60 \mathrm{~cm}$$, the length of the closed pipe will be:

Accepted Answer

$$\begin{aligned}
& \mathrm{f}_1=\frac{\mathrm{v}}{4 \mathrm{~L}_1} \
& \mathrm{f}_1=\mathrm{f}_2 \
& \frac{\mathrm{v}}{4 \mathrm{~L}_1}=\frac{\mathrm{v}}{\mathrm{L}_2} \
& \Rightarrow \mathrm{L}_2=4 \mathrm{~L}_1 \
& 60=4 	imes \mathrm{L}_1 \
& \mathrm{~L}_1=15 \mathrm{~cm}
\end{aligned}$$

Answer

15 cm

Answer

60 cm

Answer

45 cm

Answer

30 cm

Waves

The fundamental frequency of a closed organ pipe is equal to the first overtone frequency of an open organ pipe. If length of the open pipe is $60 \mathrm{~cm}$, the length of the closed pipe will be:

Choose the correct answer: