Subjects

Sections | |||
---|---|---|---|

What is a Wave? | 27 mins | 0 completed | Learn |

The Mathematical Description of a Wave | 36 mins | 0 completed | Learn |

Waves on a String | 24 mins | 0 completed | Learn |

Wave Interference | 17 mins | 0 completed | Learn |

Standing Waves | 24 mins | 0 completed | Learn |

Sound Waves | 13 mins | 0 completed | Learn |

Standing Sound Waves | 13 mins | 0 completed | Learn |

Sound Intensity | 25 mins | 0 completed | Learn |

The Doppler Effect | 17 mins | 0 completed | Learn |

Beats | 10 mins | 0 completed | Learn |

Concept #1: Standing Sound Waves

**Transcript**

Hey guys, in this video we want to talk about standing sound waves. So we're going to apply we learned about standing waves on strings to sound now. Alright, let's get to it. Just like with waves on a string when sound is reflected off of a surface, they interfere, those sound waves interfere with whatever sound waves are coming at them because they're going forward and then they're bouncing back and whatever sound waves are also coming forward there's going to be interference. Standing sound waves are produced by actual oscillations in air molecules just like regular sound waves are produced by actual oscillations in air molecules. So we have two things we need to worry about, we need to worry about the displacement of those actual air molecules and the pressure of the air. Remember the more densely packed those air molecules are the larger the pressure. It turns out that because of this a displacement node is always a pressure antinode and vice versa. A displacement antinode is always a pressure node and the reason for this is let's say that this point right here is a displacement node. What that means is those molecules are not moving but the molecules around them are oscillating back and forth. So we have these guys coming in so they're going to get real close and then they're going to move outwards so they're going to get real far so you can see that even though this line of air molecules right here is a displacement node they don't move. That area undergoes a very very large pressure change. When the two surrounding air molecules come very close and when they move very far there's low pressure. When they move very close there's high pressure, when they move very far there's low pressure so that displacement node undergoes a very very large change in pressure. Now what if this is a displacement antinode? Well then it's going to be moving back and forth and so are the molecules near it. So when this guy is moving to the left, the antinode, the molecules around it are moving with it like a wave they're all moving with it. When the displacement antinode moves the other direction all of the other ones move with it too. So those three lines sort of move as one and there is no pressure difference because the density doesn't change. They move together and because they're moving together the density isn't changing, they're not getting any closer, they're not getting any further apart so you can see that that density antinode is actually a pressure node. Node is no displacement, that pressure is not changing at that density antinode. Let me minimise myself here. Now typically with standing sound waves you were producing them inside of tubes and the open end of a tube is always a pressure node which means what? The displacement, the open end is a displacement antinode and the closed end is always a pressure antinode which means the closed end is a displacement node.

Your book and professor probably show you images that compare the two I'm going to focus entirely on pressure, I want to talk about pressure nodes and pressure antinodes alright. The two important scenarios to remember guys is a tube with both ends open has a pressure node at each end so it is a node node scenario, a tube with one end closed has a node at one end and an antinode at the other end.This is a node antinode situation besides that the equations are the same I drew figures here for a tube with two open ends and a tube with one open end and one closed end these are the pressure oscillations. So you see the pressure is a node at every open end and it's an antinode at every closed end, so the tubes on the left are node node the tubes on the right are node antinode and remember the equations for node node and node anitnode. For node node standing waves regardless of whether they're sound waves or waves on a string we have this equation for the wave lengths and this equation for the frequencies where N is any integer. Now for node antinode standing waves regardless of whether the waves on a string or sound waves right this is your equation for wavelengths allowed and this is your equation for frequencies allowed where N is only odd integers. Once again it's very very very important that you remember that is only odd integers.

Alright let me minimise myself for this problem. A wind instrument is 1 meter long and open at the mouthpiece but plugged at the far end what is the third highest harmonic frequency this instrument can make assume that the temperature is 20 degrees celsius if it's open at the mouthpiece one end but plugged at the far end it's open closed so this is node antinode. If we want to find a harmonic frequency we have to know then that the harmonic frequencies for node antinodes are N V over 4 L where N is only odd what's the third highest remember N can be 1,3,5,7 etc. What's the third highest 5, so the fifth harmonic is the third highest harmonic this 5 V over 4 L but before we can solve that we need to know the speed of sound in this instrument we're assuming that the temperatures 20 degrees so the speed is going to be 331 times 1 plus 20 over 273 three which is going to be 343 meters per second so going back to filling this out this is 5 times 343 over 4 times L. It's a one meter long instruments and this comes out to 429 hertz. Exact same equation exact same process to solving for standing sound waves just recognize are you node node or node antinode. Alright guys that wraps up our discussion on standing sound waves. Thanks for watching.

Example #1: Fundamental Frequency of Ear Canal

**Transcript**

Hey guys, let's do an example. The human ear can be modelled as a tube with one end open and one end closed. The ear canal or sorry the ear drum. If the length of the ear canal is roughly 2 and a half centimeters, what would the fundamental frequency of standing waves in the ear be? Assume that the temperature inside the ear is that of a human body, 37 degrees Celsius. If one end is open and one end is closed, this is a node-antinode problem.If we're looking for the fundamental frequency that's always a harmonic number of one. So remember any harmonic frequency for node-antinode is NV over 4L where N is odd.

The fundamental frequency is just V over 4L. Before we can solve this we need to know the speed of sound in the ear canal. We're assuming that it's at body temperature so it's the square root of 1 plus the temperature in Celsius over 273 times 331 meters per second and this equals 353 meters per second. So the fundamental frequency is 353 over 4 times the depth of the ear canal which is roughly 2 and a half centimeters or 0.025 meters which is 3530 Hertz. Just a straight up application of our equations for a node-node standing waves. Alright guys, that wraps up this problem. Thanks for watching.

Example #2: Third Harmonic for Waves in a Tube

**Transcript**

Hey guys, let's do an example. Calculate the third largest frequency for a standing sound wave in a 0.02 meter tube at 20 degrees Celsius if A, both ends are open or B, one end is closed. If both ends are open this is node-node. Remember an open end is a pressure node. If one end is closed this is a node-antinode. Remember the closed ends are pressure antinodes. So part A, FN for node-node is NV over 2L where N can be any integer. So what's the third largest? The third harmonic. N can be 1, 2 or 3.

So the third largest is the third harmonic. If the temperature is 20 degrees Celsius then the speed is going to be 331 times 1 plus 20 over 273 which is going to be 343 meters per second so the speed is, sorry, N the harmonic number's 3 that's the third largest harmonic, the speed of sound is 343 meters per second and the length is 0.02 meters and so this is 25725 Hertz. But now what if one end is closed? Now we have to use the node-antinode equations where FN is NV over 4L and what is the third largest harmonic? Well N can be 1, 3 and then 5. It has to be odd so the third largest harmonic is 5. So the fifth harmonic is 5 times 345 over 4 times 0.02 and this is 21438 Hertz. Alright, that wraps up this problem. Thanks for watching guys.

0 of 3 completed

A 0.7 m long wind instrument is open at one end and closed at the other. What is the fundamental harmonic frequency and first possible overtone frequency of the instrument?

An organ pipe of length L is open at one end and closed at the other. A tuning fork is brought near the pipe, exciting the air in the pipe with three nodes as represented in the figure.
With the speed of the sound denoted by v, what is the frequency of the tuning fork?

An organ pipe is open at both ends, with a length of 1.2 m. What are the first two overtone frequencies of the organ pipe?

The figure below represents a sound wave in a hollow pipe with both ends open. Determine the wave length of the sound wave in this hollow pipe.1. λ = 2ℓ/92. λ = ℓ3. λ = 2ℓ/74. λ = ℓ35. λ = ℓ46. λ = ℓ27. λ = 2ℓ/38. λ = 2ℓ9. λ = 2ℓ/5

Consider another organ pipe which has one end open and one end closed. Determine the wave length of the sound wave in this hollow pipe.
1. λ = 4ℓ/13
2. λ = 4ℓ/11
3. λ = 4ℓ/5
4. λ = 4ℓ/3
5. λ = 4ℓ
6. λ = 4ℓ/7
7. λ = 4ℓ/15
8. λ = 4ℓ/17
9. λ = 4ℓ/9

Consider two pipes of the same length: one pipe is open at both ends and the other pipe is closed on one end but open at the other end. If the fundamental frequency of the totally open pipe is 300 Hz, what is the fundamental frequency of the other pipe?
A) 150 Hz
B) 300 Hz
C) 450 Hz
D) 600 Hz

Using the two measured pipe lengths (L1 = 66cm and L2 = 40 cm), work out the wavelength of the sound wave. Use this to determine the mode numbers and speeds of sound that the two lengths correspond to. You can assume that L1 and L2 represent neighboring resonances (i.e, n and n+2). The pipes are open on one end and closed on the other. Frequency of tuning fork is 384 Hz.

An organ pipe is made to play a low note at 27.5 Hz, the same as the lowest note on a piano. Assuming a sound speed of 343 m/s, a. What length open-open pipe is needed? b. What length open-closed pipe would suffice?

The lowest frequency in most humans' audible range is 20 Hz.a. What is the length of the shortest open-open tube needed to produce this frequency?L = ___ mb. What is the length of the shortest open-closed tube needed to produce this frequency?L = ___ m

A piano tuner stretches a steel piano wire with a tension of 765 N. The steel wire has a length of 0.900mand a mass of 6.75 g . Part A What is the frequency f1 of the string's fundamental mode of vibration? Express your answer numerically in hertz using three significant figures. Part B What is the number n of the highest harmonic that could be heard by a person who is capable of hearing frequencies up to f = 16 kHz? Express your answer exactly.

Draw a picture of standing waves with n=3 and n=5 in a pipe with one open end. Consider the following variables: frequency f, wavelength (lamda λ), sound of speed (v), mode number n, and pipe length (L). If you used the same tuning fork to create the two standing waves, which of these variables have changed between the two standing waves and which have remained the same? Explain.

You are attempting to create a standing wave with n = 7 in a pipe .50m in length and filled with air. The top of the pipe is open to the air and the bottom of the pipe is barely submerged in water and thus closed. Draw the diagram for this standing wave with the proper nodes and antinodes for this open-closed pipe. What is the frequency?

The normal modes of this system are products of trigonometric functions. (For linear systems, the time dependance of a normal mode is always sinusoidal, but the spatial dependence need not be.) Specifically, for this system a normal mode is described byyi(x,t)=Ai sin(2p*x/?i)sin(2pfi*t)Find the three lowest normal mode frequencies f1, f2, and f3.Express the frequencies in terms of L, v, and any constants. List them in increasing order, separated by commas.

In a resonating pipe which is open at both ends, thereA) are displacement antinodes at each end.B) is a displacement node at one end and a displacement antinode at the other end.C) are displacement nodes at each end.D) none of the above

Enter your friends' email addresses to invite them:

We invited your friends!

Join **thousands** of students and gain free access to **55 hours** of Physics videos that follow the topics **your textbook** covers.