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problem 12.28 an organ pipe is 118 cm long. the speed of sound in air i…

Question

problem 12.28
an organ pipe is 118 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
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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 $L = 118\ cm=1.18\ m$. The speed of sound $v = 343\ m/s$.

Step2: Find fundamental and over - tones for closed - end pipe

For a pipe closed at one end, the fundamental frequency $f_1=\frac{v}{4L}$. Substituting values, $f_1=\frac{343}{4\times1.18}\approx72.6\ Hz$. The first overtone $f_3 = 3f_1$, the second overtone $f_5=5f_1$, and the third overtone $f_7 = 7f_1$.

  • $f_3=3\times\frac{343}{4\times1.18}\approx218\ Hz$.
  • $f_5=5\times\frac{343}{4\times1.18}\approx363\ Hz$.
  • $f_7=7\times\frac{343}{4\times1.18}\approx508\ Hz$.

Step3: Find fundamental and over - tones for open - end pipe

For a pipe open at both ends, the fundamental frequency $f_1=\frac{v}{2L}$. Substituting values, $f_1=\frac{343}{2\times1.18}\approx145\ Hz$. The first overtone $f_2 = 2f_1$, the second overtone $f_3=3f_1$, and the third overtone $f_4 = 4f_1$.

  • $f_2=2\times\frac{343}{2\times1.18}\approx290\ Hz$.
  • $f_3=3\times\frac{343}{2\times1.18}\approx434\ Hz$.
  • $f_4=4\times\frac{343}{2\times1.18}\approx579\ Hz$.

Answer:

For pipe closed at one end: $72.6,218,363,508$
For pipe open at both ends: $145,290,434,579$