# Amplitude of standing wave for higher frequency

• songoku
In summary, the amplitude of a standing wave on a string is affected by the number of nodes or frequency of the wave. As the frequency increases, the distance traveled in each cycle decreases, resulting in a decrease in amplitude. This is due to the relationship between power, pressure, and particle velocity in the wave. However, the details of how the wave is driven and loses energy can also impact the amplitude in practical situations.
songoku
Homework Statement
This is not homework

Please see below picture. In my notes, as the frequency of electric vibrator is increased, there will be more nodes produced and the amplitude will be less.
Relevant Equations
##y=2A \sin kx \cos \omega t##

I understand the part where there will be more nodes produced because number of wave produced will increase (let say from half wave to one wave). But I don't understand the part where the amplitude will be less. How can number of nodes (or frequency) affect the amplitude of standing wave produced in the string?

Thanks

The power supplied to the string by the vibration generator is proportional to pressure and particle velocity. If the power remains constant, the velocity remains constant as frequency is altered . But as frequency is increased, the distance traveled in each cycle is less because the time is less. The power in a sound wave is found by multiplying the pressure wave by the particle velocity wave. So the "amplitude" of the wave is either the pressure or the velocity but not the distance of vibration.

songoku
tech99 said:
The power supplied to the string by the vibration generator is proportional to pressure and particle velocity.
Is this the formula you mean? I found it in wikipedia

tech99 said:
If the power remains constant, the velocity remains constant as frequency is altered .
If that is the formula, then if power and velocity remain constant, then the sound pressure will also remain constant since the area will also be constant?

tech99 said:
So the "amplitude" of the wave is either the pressure or the velocity but not the distance of vibration.
Sorry I don't really understand this part.

I found something similar to my question in this link:
https://opentextbc.ca/universityphysicsv1openstax/chapter/16-6-standing-waves-and-resonance/

That is part of text in the link above. The last sentence in explanation of Figure 16.29:
"Conducting this experiment in the lab would result in a decrease in amplitude as the frequency increases."

It seems to me that the amplitude referred by the text is the vertical displacement of the particle. But I don't understand why the amplitude will decrease as the frequency increases

Thanks

songoku said:
"Conducting this experiment in the lab would result in a decrease in amplitude as the frequency increases."
This wording suggests it is a 'real world' effect rather than a theoretical one in an idealised model.
Lateral displacement will create some longitudinal displacement too. This will increase at higher frequency as that increases the path length along the string.

songoku
This is not a sound wave, it is a transverse (standing) wave on a string at constant tension. The energy in the string is easiest to calculate when it is all kinetic and $$E~\alpha~ A^2\omega^2$$ so for a fixed energy density in the wave the result is evident.
In practice the details of how the wave is driven and how it loses energy will be determinative

songoku
haruspex said:
Lateral displacement will create some longitudinal displacement too. This will increase at higher frequency as that increases the path length along the string.
Sorry I don't understand the term "lateral displacement" and "longitudinal displacement".

By lateral displacement, do you mean vertical displacement of point on the string?

hutchphd said:
This is not a sound wave, it is a transverse (standing) wave on a string at constant tension. The energy in the string is easiest to calculate when it is all kinetic and $$E~\alpha~ A^2\omega^2$$ so for a fixed energy density in the wave the result is evident.
In practice the details of how the wave is driven and how it loses energy will be determinative
I see. So I refer to the wrong formula

Thanks

songoku said:
the term "lateral displacement" and "longitudinal displacement".
Longitudinal means, in this context, in the direction of propagation of the wave. In a sound wave, all the displacement is longitudinal. In a transverse wave, most of the displacement is lateral, i.e. normal to the propagation, but there can be some longitudinal displacement too.
Anyway, hutch has pointed out the answer.

hutchphd and songoku
Thank you very much for the explanation tech99, haruspex, hutchphd

hutchphd

## 1. What is the amplitude of a standing wave for higher frequency?

The amplitude of a standing wave for higher frequency refers to the maximum displacement of the wave from its equilibrium position. In other words, it is the distance between the highest point of the wave (crest) and the lowest point of the wave (trough).

## 2. How does the amplitude of a standing wave change with higher frequency?

As the frequency of a standing wave increases, the amplitude also increases. This is because a higher frequency means that the wave is oscillating at a faster rate, resulting in a larger displacement from its equilibrium position.

## 3. What factors can affect the amplitude of a standing wave for higher frequency?

The amplitude of a standing wave for higher frequency can be affected by factors such as the tension and length of the medium through which the wave is traveling, as well as the properties of the medium itself (e.g. density, elasticity).

## 4. Can the amplitude of a standing wave for higher frequency be greater than the amplitude of a standing wave for lower frequency?

Yes, the amplitude of a standing wave for higher frequency can be greater than the amplitude of a standing wave for lower frequency. This is because the amplitude is dependent on the frequency of the wave, and a higher frequency can result in a larger amplitude.

## 5. How is the amplitude of a standing wave for higher frequency related to its energy?

The amplitude of a standing wave for higher frequency is directly proportional to its energy. This means that as the amplitude increases, so does the energy of the wave. This relationship is described by the equation E ∝ A2, where E is the energy and A is the amplitude.

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