I know that wave velocity is the product of wavelength and frequency, and that velocity is proportional to string tension. Does this mean that if you increase the tension on a string, the wavelength, frequency, and velocity will all increase?
[Physics] When you increase the tension on a string, how is the standing wave affected
forcesfrequencyoscillatorswaves
Related Solutions
Your intuition is right: the density of the string goes down a little bit when you increase the tension.
HOWEVER: the wave in a string is a transverse wave which depends on the tension and the mass per unit length. If you double the tension the mass per unit length goes down by a small amount (the string gets a bit "thinner" because it gets longer) . Both these things increase the velocity of the transverse wave which is given by
$$v = \sqrt{\frac{T}{\rho}}$$
Where T is the tension and $\rho$ the mass per unit length. Finally, the fundamental frequency is determined as the reciprocal of the round trip time of the wave along the string:
$$T = \frac{2 \ell}{v}$$
so that
$$f = \frac{1}{T} = \frac{v}{2\ell} = \frac{\sqrt{\frac{T}{\rho}}}{2\ell}$$
So to raise the frequency by an octave you need four times the tension (don't try this - you might break the neck) or a string that's half the diameter (one quarter of the area - so one quarter of the mass per unit length). This explains why different strings on the guitar have different gage - you would need too much tension to get the high range from a long thick string.
There is an equation that translate properly the situation of a string under a pulse of frequency $\omega$ at the point $x_0$ of the string. $$ \rho\frac{d^2y}{dt^2}=T\frac{d^2y}{dx^2}+\kappa \delta(x-x_0)\sin(\omega t) $$ this equation is nothing more than an application of the Newton's second law. The first term is the $ma$ part of the $f=ma$. Here $\rho$ play the hole of the mass $m$, actually is the density of the string. The logic is that each point $x$ of the string can deform into $y_x(t)$, or $y(x,t)$. $\frac{d^2y}{dt^2}$ is the acceleration of the point $x$ at time $t$. The second term is the tension.
Tension is a force that any point of the string can feel if the nearby points of the string are not stretched. $T$ measures the strongness. If you deform the sting, then the tension would try to send the string back to stretchiness but, by conservation of energy, this produces an oscillation.
The last term is the source. The $\delta(x-x_0)$ says that the source is located at point $x_0$ of the string. The $\sin(\omega t)$ says that the source oscillates the point $x_0$ with frequency $\omega$.
Putting all this together we have that as the source start to oscillate the point $x_0$, the tension start to act on other points $x\neq x_0$ close to $x_0$. This eventually happens through on all the string. The result is a wave with velocity $v=\sqrt{T/\rho}$. This is so because at points $x\neq x_0$, we have: $$ \rho\frac{d^2y}{dt^2}=T\frac{d^2y}{dx^2} \rightarrow \frac{d^2y}{dt^2}=\frac{T}{\rho}\frac{d^2y}{dx^2} \rightarrow \frac{\partial^2 y}{\partial x_+ \partial x_-}=0 $$ where $x_{\pm}=x\pm vt$. The solution is a superposition of left and right waves $$ y(x,t)= f(x_+)+g(x_-) $$ The full solution need to be compatible with the oscillation of the $x_0$ with frequency $\omega$. Then we conclude that all points need to vibrate exactly at the same frequency with different phases, otherwise, $y(x,t) \neq f(x_+)+g(x_-)$. Try to visualise by taking the referential frame that make the the configuration of the function $y(x,t)$ at reast.
Best Answer
This is something that I always see students getting tripped up with.
The equation $v=f\lambda$ just relates these variables together. It does not tell us, in general, how changing one variable changes another. You need additional information.
For example, the velocity of waves on a string is determined by $$v=\sqrt{\frac{F}{\mu}}$$ where $F$ is the tension in the string and $\mu$ is the linear mass density. The wavelength and frequency of the wave doesn't determine the speed. It's a property of the string. Also note that your statement that the velocity is proportional to the tension is incorrect.
If we are talking about generating waves on a string by wiggling one end, then the frequency is determined by how you wiggle the string. Then you can determine the wavelength of these waves by $v=f\lambda$
However, if we are talking about standing waves where only certain wavelengths are allowed, then we can determine the frequency we need to wiggle the string at by $v=f\lambda$
So, your question
does not have an answer unless you specify what system you are looking at (although the velocity increases for sure).