Question

an organ pipe is 127 cm long. the speed of sound in air is 343 m/s. what are the fundamental and first three audible overtones if the pipe is closed at one end? express your answers using three significant figures separated by commas. f1, f3, f5, f7 = hz. part b what are the fundamental and first three audible overtones if the pipe is open at both ends? express your answers using three significant figures separated by commas. f1, f2, f3, f4 = hz

Explanation:

Step1: Convert pipe length to SI unit

The length of the pipe $L = 127\ cm=1.27\ m$. The speed of sound $v = 343\ m/s$.

Step2: Calculate fundamental frequency for closed - end pipe

For a pipe closed at one end, the fundamental frequency $f_1$ is given by the formula $f_1=\frac{v}{4L}$. Substituting $v = 343\ m/s$ and $L = 1.27\ m$, we get $f_1=\frac{343}{4\times1.27}\ Hz\approx67.6\ Hz$.

Step3: Calculate first three overtones for closed - end pipe

The overtones for a pipe closed at one end are odd - multiples of the fundamental frequency. So, $f_3 = 3f_1$, $f_5=5f_1$ and $f_7 = 7f_1$.
$f_3=3\times\frac{343}{4\times1.27}\ Hz\approx203\ Hz$
$f_5 = 5\times\frac{343}{4\times1.27}\ Hz\approx338\ Hz$
$f_7=7\times\frac{343}{4\times1.27}\ Hz\approx473\ Hz$

Step4: Calculate fundamental frequency for open - end pipe

For a pipe open at both ends, the fundamental frequency $f_1=\frac{v}{2L}$. Substituting $v = 343\ m/s$ and $L = 1.27\ m$, we get $f_1=\frac{343}{2\times1.27}\ Hz\approx135\ Hz$.

Step5: Calculate first three overtones for open - end pipe

The overtones for a pipe open at both ends are integer - multiples of the fundamental frequency. So, $f_2 = 2f_1$, $f_3=3f_1$ and $f_4 = 4f_1$.
$f_2=2\times\frac{343}{2\times1.27}\ Hz\approx270\ Hz$
$f_3=3\times\frac{343}{2\times1.27}\ Hz\approx404\ Hz$
$f_4=4\times\frac{343}{2\times1.27}\ Hz\approx539\ Hz$

Answer:

For pipe closed at one end: $67.6,203,338,473$
For pipe open at both ends: $135,270,404,539$