Standing waves on pipes IB DP Physics Study Notes - 2025 Syllabus
Standing waves on pipes IB DP Physics Study Notes
Standing waves on pipes IB DP Physics Study Notes at IITian Academy focus on specific topic and type of questions asked in actual exam. Study Notes focus on IB Physics syllabus with Students should understand
standing waves patterns in strings and pipes
Standard level and higher level: 4 hours
Additional higher level: There is no additional higher level content .
- IB DP Physics 2025 SL- IB Style Practice Questions with Answer-Topic Wise-Paper 1
- IB DP Physics 2025 HL- IB Style Practice Questions with Answer-Topic Wise-Paper 1
- IB DP Physics 2025 SL- IB Style Practice Questions with Answer-Topic Wise-Paper 2
- IB DP Physics 2025 HL- IB Style Practice Questions with Answer-Topic Wise-Paper 2
Boundary conditions – closed pipes
- We can also set up standing waves in pipes.
- In the case of pipes, longitudinal waves are created (instead of transverse waves), and these waves are reflected from the ends of the pipe.
- Consider a closed pipe of length L which gets its wave energy from a mouthpiece on the left side.
Boundary conditions – open pipes
- In an open-ended pipe you have an antinode at the open end because the medium can vibrate there (and, of course, another antinode at the mouthpiece).
Distinguishing between standing and traveling waves
Example:
An open pipe and a closed pipe have the same length of \( 0.85 \, \text{m} \). The speed of sound in air is \( 340 \, \text{m/s} \).
(a) Calculate the fundamental frequency of each pipe.
(b) Which pipe will have the higher pitch and why?
▶️ Answer/Explanation
(a) Fundamental Frequencies:
For an open pipe (both ends open), the fundamental wavelength is \( \lambda = 2L = 2 \times 0.85 = 1.70 \, \text{m} \)
So, \( f = \frac{v}{\lambda} = \frac{340}{1.70} = \boxed{200 \, \text{Hz}} \)
For a closed pipe (one end closed), the fundamental wavelength is \( \lambda = 4L = 4 \times 0.85 = 3.40 \, \text{m} \)
So, \( f = \frac{v}{\lambda} = \frac{340}{3.40} = \boxed{100 \, \text{Hz}} \)
(b) Which has the higher pitch?
The open pipe has a higher fundamental frequency (200 Hz vs 100 Hz).
Therefore, the open pipe produces a higher pitch.
IB Physics Standing waves on pipes Exam Style Worked Out Questions
Question
A pipe containing air is closed at one end and open at the other. The third harmonic standing wave for this pipe has a frequency of \(150 \mathrm{~Hz}\).
What other frequency is possible for a standing wave in this pipe?
A. \(25 \mathrm{~Hz}\)
B. \(50 \mathrm{~Hz}\)
C. \(75 \mathrm{~Hz}\)
D. \(300 \mathrm{~Hz}\)
▶️Answer/Explanation
Ans:B
In a closed-open pipe (like an open-end organ pipe), the fundamental frequency (first harmonic) is produced when the length of the pipe is one-fourth (1/4) of the wavelength of the sound wave. In this case, the pipe is closed at one end and open at the other.
The third harmonic has a frequency of \(150 \, \text{Hz}\), which means that the pipe length corresponds to one and a half wavelengths (\(\lambda/2\)). The fundamental frequency (\(f_1\)) corresponds to a quarter-wavelength (\(\lambda/4\)).
So, if the third harmonic is at \(150 \, \text{Hz}\), we can find the fundamental frequency (\(f_1\)) as follows:
\(\frac{f_3}{f_1} = \frac{\lambda_3}{\lambda_1} = \frac{3}{1}\)
\(f_1 = \frac{f_3}{3} = \frac{150 \, \text{Hz}}{3} = 50 \, \text{Hz}\)
Therefore, the possible frequency for a standing wave in this pipe, other than the third harmonic, is the fundamental frequency, which is \(50 \, \text{Hz\).
Question
A pipe of length 0.6 m is filled with a gas and closed at one end. The speed of sound in the gas is 300 m s–1. What are the frequencies of the first two harmonics in the tube?
A 125 Hz and 250 Hz
B 125 Hz and 375 Hz
C 250 Hz and 500 Hz
D 250 Hz and 750 Hz
Answer/Explanation
ANSWER – B
A pipe of length L with one open end produces the longest wavelength (λ) (i.e., lowest frequency) which is 4 times the length of the pipe. That is:
\(\Lambda=4L\)
Therefore, the lowest/fundamental frequency (f), at speed of sound (v) is:
\(f=\frac{v}{\Lambda }=\frac{v}{4L}Hz\)
Further, a pipe with one open end only supports odd harmonics. So, the second harmonic f(2) is not produced. The next harmonic which is produced is the third harmonic f(3) which is 3 times the fundamental frequency:
\(f\left ( 3 \right )=3\times \frac{v}{4L}Hz\)
Substitute values from the problem statement:
L = 0.6 meters
\(v=300ms^{-1}\)