Velocity & Frequency Wave on String

[escalade17]

Please help me, I'm so bewildered it's not even funny.

Does the frequency only depend on the source? This all seems so paradoxical to me...for example:

An 100 Hz oscillator produces a wave on a rope with a wavelength of .1 meters. If I increase the tension of the rope by sticking a weight at the end of it, the velocity increases. So does the wavelength, but the frequency stays he same.

But...if I have a standing wave on a string and I increase the tension of the string, the frequency of the wave definitely increases. This frequency given as f=nv/2L (because the velocity increased the frequency does too).

Is this all because in the first example the oscillator produces a constant frequency? If the oscillator had not been there, wouldn't the frequency increase?

Also, if I take the first example and turn the oscillator to 200 Hz, wouldn't the wavelength drop correspondingly but the velocity of the wave would stay constant (velocity only determined by (Tension/linear mass density)^1/2.

 

[mfb]

If you do not have reflections at the other end, you get an oscillation with a frequency equal to the oscillator frequency. With reflections, it can get tricky to define "frequency" at all. You can always write the whole system as superposition of resonance frequencies, but then you do not have "one frequency", but many.

Is this all because in the first example the oscillator produces a constant frequency? If the oscillator had not been there, wouldn't the frequency increase?

I do not understand which setup you have in mind here.

Also, if I take the first example and turn the oscillator to 200 Hz, wouldn't the wavelength drop correspondingly but the velocity of the wave would stay constant

Correct.

 

[escalade17]

Thank you mfb.

I guess I'm talking about something similar to what they have set up here:

http://www.niiler.com/phy130/lab11waves.pdf

Now, the only way to increase the speed of that wave would be to put a bigger mass on the other end, right? But what happens to the frequency of the wave if I do that. Wouldn't it stay the same because the oscillator is directing the frequency? Normally as the velocity of the wave increases the frequency does too (freq: nv/2L), but does the oscillator take prescience?
So the only things that should happen would be the freq of that wave increases and correspondingly the wavelength (but the frequency remains at the frequency set by the oscillator).

 

[sophiecentaur]

We just had a long thread about this same basic thing. (Here)
Ignoring the transitory effect on an oscillating system when an excitation is initially switched on (in which some of the natural modes can be excited briefly by a step function, but then die out), the only vibrations that can exist can be at the excitation frequency. Depending upon how close the exciting frequency is to a natural mode of vibration and the Q factor of the oscillator, the standing wave will be at a high or low level.
There is no way that such a linear system can produce a change of frequency because you need to have phase continuity with the energy supply and a change in frequency would violate this.

 

[PJC01]

I can understand how this may seem confusing and paradoxical. However, there are some key concepts that can help clarify this issue.

Firstly, frequency is defined as the number of cycles or oscillations per unit of time. In the first example, the frequency is determined by the oscillator and is constant regardless of the tension or velocity of the wave on the rope. This is because the oscillator is the source of the wave and is responsible for determining its frequency.

In the second example, the frequency of the standing wave on the string is determined by the tension and velocity of the wave. As you correctly pointed out, the equation f=nv/2L shows that an increase in velocity results in an increase in frequency. This is because the standing wave is not produced by an external source, but rather is a result of the interaction between the string and the tension applied to it.

So, to answer your question, yes, the frequency in the first example is only dependent on the source (the oscillator), while in the second example, it is dependent on both the source (the tension) and the velocity of the wave.

As for your question about the wavelength, it is important to note that wavelength and frequency are inversely proportional. This means that as one increases, the other decreases. In the second example, when the tension is increased, the wavelength decreases, resulting in an increase in frequency. This is because the standing wave is now able to fit more cycles within the shorter wavelength.

In summary, the key difference between the two examples is the source of the wave. In the first example, the frequency is determined by the source (oscillator), while in the second example, it is determined by the interaction between the string and the tension. I hope this helps clarify the concept of velocity and frequency in waves on a string.