Amplitude of standing wave for higher frequency

[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

 

[tech99]

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]

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

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?

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

 

[haruspex]

"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.

 

[hutchphd]

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]

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?

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

 

[haruspex]

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.

 

[songoku]

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