Harmonics and Standing Waves - AP Physics 1

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Question

Which of the following is a standing wave?

Answer

A simple definition of a standing wave is a wave that is self-reinforcing, which is to say that reflection of the wave through the medium results in some areas of amplification (anti-nodes) of the wave and some areas of nullification (nodes). In other words, resonance must occur, and that usually suggests confinement of the wave in some fashion.

A fan and a bus make noise and vibration, but the sound does not resonate. It is transmitted, but not confined. Light with a specific wavelength has no "resonant" character, and neither do waves striking a pier. If the waves were confined in a harbor so that they could amplify, it might be possible to produce a standing wave. Microwaves trapped inside a microwave oven have this feature, producing antinodes of intense heating and nodes where no energy is transmitted into the food; this is the reason that microwave ovens have rotating platforms to make heating of the food item more uniform.

A violin string will be seen to have discrete, stable regions of motion and lack of motion, the requirements of the standing wave phenomenon. The points of reflection on the string are the two ends. The vibration of the wave is confined within the string, amplifying the sound as the nodes overlap.

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Question

A guitar string has a length of . If the string is vibrating with a wavelength of , what harmonic is it vibrating at?

Answer

Guitar strings are attached to the guitar at both ends; therefore, each end of the string is a node. From this, we can say that the first harmonic contains only a single antinode. Each time we add another antinode and node (or half of a wave), we reach the next harmonic. We can say that the number of antinodes in the string tells us what harmonic is being played.

The problem statement tells us that the string length is 0.5m and the wavelengths are 0.25meters. This tells us there are two complete waves in the guitar string. This gives us a total of four antinodes; thus we are in the fourth harmonic.

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Question

A string of length held at a tension and attached to a frequency generator oscillating at frequency is set up such that a standing wave is seen. The tension is then adjusted. Which one of these new tensions will exhibit a standing wave in the system?

Answer

The equation for frequency of a standing wave on a string is:

$f = \frac{n}{2L}\sqrt{\frac{T}{\mu}}$

This holds true if takes integer values. When adjusting the value of to the answer choices, only will maintain this equality.

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Question

Given a string of length 2m with two fixed ends, what is the longest wavelength of a standing wave that is possible?

Answer

The wavelengths possible for a standing wave on a string with 2 open ends are:

$\lambda=\frac{2L}{n}$

Where is the length of the string and is the harmonic given in integers The longest wavelength possible for standing wave occurs when , therefore:

$\lambda=2L=4m$

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Question

Given an open pipe of length , what is its fundamental frequency?

Assume the wave passing through the pipe is a sound wave with velocity :

$v=322 \frac{m}{s}$

Answer

Firstly, since this is an open pipe, the equation for all of its harmonics based on wavelength can be given as:

$\lambda=\frac{2L}{n}$ , where is the length of the column, $\lambda$ is the wavelength of the wave, and is the harmonic given in integers.

Using the relationship between wavelength $\lambda$ and frequency :

$f=\frac{v}{\lambda}= \frac{vn}{2L}$ , where is the wave speed.

Since we are talking about sound waves in air, we know that its wave velocity is:

$v=322 \frac{m}{s}$ . We also know the column length is

We also know that the fundamental frequency occurs when . Therefore:

$f=\frac{nv}{2L}=\frac{1\cdot322 \frac{m}{s}}{2\cdot4m}=40.25s^{-1}=40.25Hz$

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Question

Some students are trying to determine the depth of a well. They drop a stone from rest and time the fall to the well's bottom. They find the time to be $\Delta t=0.9:s$ with an experimental uncertainty of $\pm 10 ; percent$ . Since they need to know the depth with more precision, they create a pure sound tone at the top of the well and note a resonance when the tone has a frequency of . The speed of sound on that day was $v=342\frac{m}{s}$ . How deep was the well?

Answer

We start by finding the range of depths allowed by the students' first experiment:

Find the largest time that's within their experimental error:

$\Delta t=0.9s + 0.09s=0.99s$

and the shortest time:

$\Delta t=0.9s - 0.09s=0.81s$

Use these to find the maximum and minimum depths using the kinematics equation:

$\Delta y=\frac{1}{2}g\Delta t^2=\frac{1}{2}\times 9.8\frac{m}{s^2}\times (0.99s)^2=4.8m$ is the maximum and

$\Delta y=\frac{1}{2}g\Delta t^2=\frac{1}{2}\times 9.8\frac{m}{s^2}\times (0.81s)^2=3.2m$

Now we have to use the sound information to get a more precise answer. The well is open at one end and closed at the other, so the resonant wavelengths are given by:

$L=\frac{\lambda}{4}+m\frac{\lambda}{2} ;;where ;m=0,1,2,3...$

Since the fundamental resonance is a quarter of the wavelength. Find $\lambda$ using the wave equation:

$v=f\lambda \rightarrow \lambda=\frac{v}{f}=\frac{305\frac{m}{s}}{285Hz}=1.2;m$

We don't know which of the harmonics, or overtones the students were hearing, so we try the integers until we find a resonant length between and . For :

$L=\frac{\lambda}{4}+m\frac{\lambda}{2}=\frac{1.2m}{4}+0(\frac{1.2m}{2})=0.3 m$

Which is way too small. Try the other integers:

$L=\frac{\lambda}{4}+m\frac{\lambda}{2}=\frac{1.2m}{4}+1(\frac{1.2m}{2})=1.5 m$

$L=\frac{\lambda}{4}+m\frac{\lambda}{2}=\frac{1.2m}{4}+2(\frac{1.2m}{2})=2.7 m$

$L=\frac{\lambda}{4}+m\frac{\lambda}{2}=\frac{1.2m}{4}+3(\frac{1.2m}{2})=3.9 m$

$L=\frac{\lambda}{4}+m\frac{\lambda}{2}=\frac{1.2m}{4}+4(\frac{1.2m}{2})=5.1 m$

Any of these would resonate, but the only one that's within the students' kinematics margin of error corresponds to , so the well must be deep.

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Question

What is the fundamental frequency of a standing wave traveling at a speed of $20 \frac{m}{s}$ through a string of length ?

Answer

Frequency of a standing wave is related to the wave speed and length by:

$f=\frac{vn}{2L}$ , where is wave speed, and is the length of the material, and is the harmonic, given in integers.

Since the fundamental frequency is given when ,

Plugging in $v=20 \frac{m}{s}$ , and

$f=\frac{20 \frac{m}{s}}{2*4}=2.5 Hz$

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Question

A standing wave occurs within a string of length . What will be the wavelength of the third harmonic?

Answer

Recall that each additional harmonic will increase the frequency by a factor of $\frac{n}{2}$ , where is the harmonic number. Conversely, the wavelength will decrease by a factor of $\frac{2}{n}$ . Since we are on the third harmonic , and the length of the string is , the wavelength of the standing wave will be:

$\frac{2}{3}*3m= 2m$

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Question

Strings satisfy an important equation known as the wave equation. The solution to the wave equation of a point on the string over time can be given as:

$y(t)=\sum_{n=1}^{\infty}sin(\frac{n\pi t}{l})$ , where is the harmonic, is the length of the string.

Determine the period of the fundamental frequency.

Answer

Recall that for sinusoids of form , the period is given by:

$period=\frac{2\pi}{b}$

For this problem, the fundamental frequency is when , which means that its frequency is:

$period=\frac{2\pi}{\frac{\pi}{l}}=2l$

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Question

What is the wavelength produced on the third harmonic of a long open pipe?

Answer

The formula for wavelength depending on the harmonic is as follows:

$\lambda = \frac{2L}{n}$

Where $\lambda$ is the wavelength, is the length of the pipe and n is the harmonic number. Substituting our values in the equation we obtain:

$\lambda = \frac{2(4m)}{3} = \frac{8}{3} = 2\frac{2}{3}m$

Therefore the correct answer is $2 \frac{2}{3}m$ long.

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Question

Wave 1 has an amplitude of .

Wave 2 has an amplitude of .

What is the maximum and minimum amplitude of these waves when they undergo interference with each other?

Answer

The correct answer is and .

This is because at both waves maxima they would add constructively in the form of in the positive axis.

When they interfere destructively they would subtract in the form of in the positive axis.

Therefore the answer is and .

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Question

Suppose that a pipe open at one end supports a wave whose wavelength is . If this wave is on its third harmonic, how long is the pipe?

Answer

To answer this question, we need to understand the concept of standing waves. A standing wave is a wave that exhibits both nodes and antinodes. A node is a fixed position where there is no displacement from the wave, and an antinode is a point of maximum displacement of the wave.

A standing wave can result from two individual waves traveling in opposite directions and interfering with each other. For instance, when a wave traveling through one medium (such as the air in the pipe) hits an interface that is denser (such as the end of the pipe), the wave will reflect back. However, this reflected wave will be shifted $180^{o}$ out of phase. As a result, the reflected wave will interact with the incident wave, and the two will interfere. This interference takes on a specific pattern in which there is destructive interference in the location of nodes, and constructive interference in the location of the antinodes. In such a situation, it doesn't appear that the wave is traveling at all, but rather it looks as if it's standing still, hence the name standing wave.

To solve this question, we need to use the expression for a pipe closed at one end and open at the other.

$L=n\frac{\lambda }{4}$

With this expression, we just need to plug in the values given to us in the question to solve for the length of the pipe.

$L=3(\frac{440: cm}{4})=330:cm$

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Question

Which of the following is a standing wave?

Answer

A simple definition of a standing wave is a wave that is self-reinforcing, which is to say that reflection of the wave through the medium results in some areas of amplification (anti-nodes) of the wave and some areas of nullification (nodes). In other words, resonance must occur, and that usually suggests confinement of the wave in some fashion.

A fan and a bus make noise and vibration, but the sound does not resonate. It is transmitted, but not confined. Light with a specific wavelength has no "resonant" character, and neither do waves striking a pier. If the waves were confined in a harbor so that they could amplify, it might be possible to produce a standing wave. Microwaves trapped inside a microwave oven have this feature, producing antinodes of intense heating and nodes where no energy is transmitted into the food; this is the reason that microwave ovens have rotating platforms to make heating of the food item more uniform.

A violin string will be seen to have discrete, stable regions of motion and lack of motion, the requirements of the standing wave phenomenon. The points of reflection on the string are the two ends. The vibration of the wave is confined within the string, amplifying the sound as the nodes overlap.

Compare your answer with the correct one above

Question

A guitar string has a length of . If the string is vibrating with a wavelength of , what harmonic is it vibrating at?

Answer

Guitar strings are attached to the guitar at both ends; therefore, each end of the string is a node. From this, we can say that the first harmonic contains only a single antinode. Each time we add another antinode and node (or half of a wave), we reach the next harmonic. We can say that the number of antinodes in the string tells us what harmonic is being played.

The problem statement tells us that the string length is 0.5m and the wavelengths are 0.25meters. This tells us there are two complete waves in the guitar string. This gives us a total of four antinodes; thus we are in the fourth harmonic.

Compare your answer with the correct one above

Question

A string of length held at a tension and attached to a frequency generator oscillating at frequency is set up such that a standing wave is seen. The tension is then adjusted. Which one of these new tensions will exhibit a standing wave in the system?

Answer

The equation for frequency of a standing wave on a string is:

$f = \frac{n}{2L}\sqrt{\frac{T}{\mu}}$

This holds true if takes integer values. When adjusting the value of to the answer choices, only will maintain this equality.

Compare your answer with the correct one above

Question

Given a string of length 2m with two fixed ends, what is the longest wavelength of a standing wave that is possible?

Answer

The wavelengths possible for a standing wave on a string with 2 open ends are:

$\lambda=\frac{2L}{n}$

Where is the length of the string and is the harmonic given in integers The longest wavelength possible for standing wave occurs when , therefore:

$\lambda=2L=4m$

Compare your answer with the correct one above

Question

Given an open pipe of length , what is its fundamental frequency?

Assume the wave passing through the pipe is a sound wave with velocity :

$v=322 \frac{m}{s}$

Answer

Firstly, since this is an open pipe, the equation for all of its harmonics based on wavelength can be given as:

$\lambda=\frac{2L}{n}$ , where is the length of the column, $\lambda$ is the wavelength of the wave, and is the harmonic given in integers.

Using the relationship between wavelength $\lambda$ and frequency :

$f=\frac{v}{\lambda}= \frac{vn}{2L}$ , where is the wave speed.

Since we are talking about sound waves in air, we know that its wave velocity is:

$v=322 \frac{m}{s}$ . We also know the column length is

We also know that the fundamental frequency occurs when . Therefore:

$f=\frac{nv}{2L}=\frac{1\cdot322 \frac{m}{s}}{2\cdot4m}=40.25s^{-1}=40.25Hz$

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Question

Some students are trying to determine the depth of a well. They drop a stone from rest and time the fall to the well's bottom. They find the time to be $\Delta t=0.9:s$ with an experimental uncertainty of $\pm 10 ; percent$ . Since they need to know the depth with more precision, they create a pure sound tone at the top of the well and note a resonance when the tone has a frequency of . The speed of sound on that day was $v=342\frac{m}{s}$ . How deep was the well?

Answer

We start by finding the range of depths allowed by the students' first experiment:

Find the largest time that's within their experimental error:

$\Delta t=0.9s + 0.09s=0.99s$

and the shortest time:

$\Delta t=0.9s - 0.09s=0.81s$

Use these to find the maximum and minimum depths using the kinematics equation:

$\Delta y=\frac{1}{2}g\Delta t^2=\frac{1}{2}\times 9.8\frac{m}{s^2}\times (0.99s)^2=4.8m$ is the maximum and

$\Delta y=\frac{1}{2}g\Delta t^2=\frac{1}{2}\times 9.8\frac{m}{s^2}\times (0.81s)^2=3.2m$

Now we have to use the sound information to get a more precise answer. The well is open at one end and closed at the other, so the resonant wavelengths are given by:

$L=\frac{\lambda}{4}+m\frac{\lambda}{2} ;;where ;m=0,1,2,3...$

Since the fundamental resonance is a quarter of the wavelength. Find $\lambda$ using the wave equation:

$v=f\lambda \rightarrow \lambda=\frac{v}{f}=\frac{305\frac{m}{s}}{285Hz}=1.2;m$

We don't know which of the harmonics, or overtones the students were hearing, so we try the integers until we find a resonant length between and . For :

$L=\frac{\lambda}{4}+m\frac{\lambda}{2}=\frac{1.2m}{4}+0(\frac{1.2m}{2})=0.3 m$

Which is way too small. Try the other integers:

$L=\frac{\lambda}{4}+m\frac{\lambda}{2}=\frac{1.2m}{4}+1(\frac{1.2m}{2})=1.5 m$

$L=\frac{\lambda}{4}+m\frac{\lambda}{2}=\frac{1.2m}{4}+2(\frac{1.2m}{2})=2.7 m$

$L=\frac{\lambda}{4}+m\frac{\lambda}{2}=\frac{1.2m}{4}+3(\frac{1.2m}{2})=3.9 m$

$L=\frac{\lambda}{4}+m\frac{\lambda}{2}=\frac{1.2m}{4}+4(\frac{1.2m}{2})=5.1 m$

Any of these would resonate, but the only one that's within the students' kinematics margin of error corresponds to , so the well must be deep.

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Question

What is the fundamental frequency of a standing wave traveling at a speed of $20 \frac{m}{s}$ through a string of length ?

Answer

Frequency of a standing wave is related to the wave speed and length by:

$f=\frac{vn}{2L}$ , where is wave speed, and is the length of the material, and is the harmonic, given in integers.

Since the fundamental frequency is given when ,

Plugging in $v=20 \frac{m}{s}$ , and

$f=\frac{20 \frac{m}{s}}{2*4}=2.5 Hz$

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Question

A standing wave occurs within a string of length . What will be the wavelength of the third harmonic?

Answer

Recall that each additional harmonic will increase the frequency by a factor of $\frac{n}{2}$ , where is the harmonic number. Conversely, the wavelength will decrease by a factor of $\frac{2}{n}$ . Since we are on the third harmonic , and the length of the string is , the wavelength of the standing wave will be:

$\frac{2}{3}*3m= 2m$

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