Answer
a. 0.585 m
b. 858 Hz
Work Step by Step
See figure 12-11. For a pipe open at both ends, the fundamental frequency is $f_1=\frac{v}{2 \mathcal{l}}$.
a. At that temperature, the speed of sound is 344.2 m/s, using the formula on page 329.
$$\mathcal{l}=\frac{v}{2f_1}=\frac{344.2m/s}{2(294Hz)}=0.585m $$
b. Look up the speed of sound in helium in Table 12–1: it is 1005 m/s. Find the pipe’s fundamental frequency.
$$f_1=\frac{1005 m/s}{2 (0.58537m)}=858.4Hz \approx 858 Hz$$