A small misunderstanding about waves

[NooDota]

Homework Statement

Hello, I've recently started learning about waves, and there are a few stuff I don't understand (I'm talking about transverse waves):1. What function describes the movement of a single "element" on the string? I mean an element in which only its y changes, but x remains constant. Its graph would be abstract, correct? What describes it?

2. What function describes the wave itself?I read that y(x,t)=h(kx-wt) describes ANY wave, where h is ANY function, is this correct? Isn't it missing Ymax?

So y(x,t)=Sin(kx-wt) is a wave function?

So is y(x,t)=x^2(kx-wt)?

So is y(x,t)=ln(kx-wt)?

2. The speed of a wave along a stretched ideal string depends only on the tension and linear density of the string and not on the frequency of the wave.

How can this be true if V= λ*ƒ?

Thank you.

Homework Equations

The Attempt at a Solution

 

[mfb]

1. What function describes the movement of a single "element" on the string? I mean an element in which only its y changes, but x remains constant. Its graph would be abstract, correct? What describes it?

For a wave with a single frequency, it follows a sine curve (as function of time).

I read that y(x,t)=h(kx-wt) describes ANY wave, where h is ANY function, is this correct? Isn't it missing Ymax?

If h is an arbitrary function, it can include an amplitude.

So y(x,t)=Sin(kx-wt) is a wave function?

Yes.

So is y(x,t)=x^2(kx-wt)?

No. It cannot depend on x like that.

So is y(x,t)=ln(kx-wt)?

You probably want some function that allows to consider negative values. Apart from that, yes.

How can this be true if V= λ*ƒ?

Where is the problem? You cannot change one without also changing the other.

 

[NooDota]

> No. It cannot depend on x like that.

Why not? "x" is a function. It has x and t, so it's a wave function, no?

 

[mfb]

"x" is a function.

It is not.

It has x and t, so it's a wave function, no?

You cannot have an arbitrary dependence, at least not if you want to describe a wave traveling along a string (or some other nondispersive homogeneous[/size] medium). It can depend on "kx-wt" only, not on x or t individually.

 

[Herman Trivilino]

Homework Statement

2. The speed of a wave along a stretched ideal string depends only on the tension and linear density of the string and not on the frequency of the wave.

How can this be true if V= λ*ƒ?

Take a string and stretch it so it's tight. Making the usual assumptions (what they refer to as the "ideal" model) the speed is fixed because the tension and density are fixed. Therefore the product $\lambda f$ is fixed. If you increase $f$ the $\lambda$ will be smaller, etc.

 

[Herman Trivilino]

Homework Statement

1. What function describes the movement of a single "element" on the string? I mean an element in which only its y changes, but x remains constant. Its graph would be abstract, correct? What describes it?

Consider a wave equation of the form $y=A \sin(kx-\omega t)$.

Choose a fixed value of $x$ and you get the equation that describes simple harmonic motion.

$y=A \sin(\omega t+\phi)$.

The graph of $y$ versus $x$ would be a single point whose $x$ coordinate was already chosen, and whose $y$ coordinate would depend on the value of $t$.

 

[Ray Vickson]

> No. It cannot depend on x like that.

Why not? "x" is a function. It has x and t, so it's a wave function, no?

If (as you wrote) you mean $y(x,t) = x^2 \cdot (kx - wt)$, then NO, that cannot be a wave function---it is not a solution of the "wave equation". However, if you were just inventing some weird notation for y(x,t) = SQUARE(kx - wt), then YES, that could be a wave function. If that is what you meant, you should have written y(x,t) = (kx - wt)^2, or $y(x,t) = (kk - wt)^2$.

 

[NooDota]

It is not.You cannot have an arbitrary dependence, at least not if you want to describe a wave traveling along a string (or some other nondispersive homogeneous medium). It can depend on "kx-wt" only, not on x or t individually.

How is x not a function? F(x)=x?

Take a string and stretch it so it's tight. Making the usual assumptions (what they refer to as the "ideal" model) the speed is fixed because the tension and density are fixed. Therefore the product $\lambda f$ is fixed. If you increase $f$ the $\lambda$ will be smaller, etc.

Yeah, but what's the relation between lambda and the period? K=2pi/lambda and W=2pi/period, I don't see the relation between the two.

Consider a wave equation of the form $y=A \sin(kx-\omega t)$.

Choose a fixed value of $x$ and you get the equation that describes simple harmonic motion.

$y=A \sin(\omega t+\phi)$.

The graph of $y$ versus $x$ would be a single point whose $x$ coordinate was already chosen, and whose $y$ coordinate would depend on the value of $t$.

Thank you.

If (as you wrote) you mean $y(x,t) = x^2 \cdot (kx - wt)$, then NO, that cannot be a wave function---it is not a solution of the "wave equation". However, if you were just inventing some weird notation for y(x,t) = SQUARE(kx - wt), then YES, that could be a wave function. If that is what you meant, you should have written y(x,t) = (kx - wt)^2, or $y(x,t) = (kk - wt)^2$.

Consider now a wave of arbitrary shape, given by
y(x, t)= h(kx+- wt)
where h represents any function, the sine function being one possibility. Our
previous analysis shows that all waves in which the variables x and t enter
into the combination kx+- wt are traveling waves. Furthermore, all traveling
waves must be of the form of Eq. 16-17. Thus, y(x, t) represents a
possible (though perhaps physically a little bizarre) traveling wave. The function
y(x, t) = sin(ax^2+- bt), on the other hand, does not represent a traveling wave.The book says h is any function, it doesn't specify anything about it, as long as it has ax+-bt in it.

 

[Ray Vickson]

How is x not a function? F(x)=x?

Yeah, but what's the relation between lambda and the period? K=2pi/lambda and W=2pi/period, I don't see the relation between the two.

Thank you.
Consider now a wave of arbitrary shape, given by
y(x, t)= h(kx+- wt)
where h represents any function, the sine function being one possibility. Our
previous analysis shows that all waves in which the variables x and t enter
into the combination kx+- wt are traveling waves. Furthermore, all traveling
waves must be of the form of Eq. 16-17. Thus, y(x, t) represents a
possible (though perhaps physically a little bizarre) traveling wave. The function
y(x, t) = sin(ax^2+- bt), on the other hand, does not represent a traveling wave.The book says h is any function, it doesn't specify anything about it, as long as it has ax+-bt in it.

You ask about the relationship between k and w. That depends on underlying properties of the transmission medium, through the wave speed. This will be different for waves in a taut string, for water waves, sound waves, light waves, etc.

 

[mfb]

How is x not a function? F(x)=x?

x is like 5. It is a value.
If you define F(x)=x (and define domain and codomain)[/size], then F is a function. Not x, not F(x) (those are both values), but F is a function.

Yeah, but what's the relation between lambda and the period?

$v=f \lambda$ is the relation. You can also write it as $f=\frac{v}{\lambda}$ or $\lambda=\frac{v}{f}$, all three equations are equivalent. $T=\frac 1 f$ which you can replace in any equation. As an example, $\lambda=v T$

The book says h is any function, it doesn't specify anything about it, as long as it has ax+-bt in it.

ax+-bt and nothing else that would depend on x or t.

 

[Herman Trivilino]

Yeah, but what's the relation between lambda and the period? K=2pi/lambda and W=2pi/period, I don't see the relation between the two.

It's the wave speed. $\lambda =vT$. Also, $v=\omega k$.

The book says h is any function, it doesn't specify anything about it, as long as it has ax+-bt in it.

No, it doesn't say it's any function. It says it's any function of $kx \pm \omega t$.

 

[NooDota]

Okay, thank you.

So increasing the frequency leads to increasing the tension forces?

 

[Herman Trivilino]

So increasing the frequency leads to increasing the tension forces?

Suppose, for example, you are tuning a guitar string. The wavelength stays the same as you increase the tension, resulting in a higher frequency.

If you had a vibrating string and all you did was increase the frequency, it wouldn't increase the tension.

The important thing to understand here is that the wave speed depends only on properties of the medium. Thus the frequency and wavelength are inversely proportional. This occurs in many situations, such as sound waves.

 

[mfb]

Okay, thank you.

So increasing the frequency leads to increasing the tension forces?

If your wavelength is fixed, like for a guitar string, then the two are correlated like that. The causal relationship in usual setups is in the opposite direction - you fix the tension, and the frequency is then given by the tension (and other constants).