Answer:
460 Hz
Step-by-step explanation:
the given resonating frequency of the device
f₁ = 920 Hz and f₂ = 1380 Hz
fundamental frequency of the device is
f₀ = n₂ - n₁
= 1380 - 920
= 460 Hz
expression of frequency of organic pipe open at both ends
![f_0=n(\\u)/(2l)](https://img.qammunity.org/2020/formulas/physics/college/f40akg7qk21gnicdwukwi6du8d9xg3pizi.png)
at n = 1
![f_0=(\\u)/(2l) = 460 Hz](https://img.qammunity.org/2020/formulas/physics/college/snck2b3efmsnm6vv53cbeg89nuxdg7rf8g.png)
the frequency ratios of the closed pipe
![f_0:f_1:f_2: ...... =[1:2:3:.........]f_0](https://img.qammunity.org/2020/formulas/physics/college/8f86tgkn38ymz4q147974x760lhxxtryio.png)
=[1:2:3:.........]460 Hz
= 460 Hz : 920 Hz : 1380 Hz
so, the lowest frequency for the pipe open at both end is 460 Hz