Frequency and wavelength of a wave

[Shay10825]

Hi everyone .

If I had a sine wave on a string at one instant of time how come when I reduce the wavelength and frequency by half the graph of the wave does not change.

~Thanks

 

[Gokul43201]

What are you plotting on the graph ?

 

[Shay10825]

The question said :

The figure below shows a sine wave on a string at one instant of time (it is a normal sine wave). Which of the graphs below shows a wave where the wavelength and frequency asr each reduced by half? The answer was the same graph.

 

[futb0l]

Well, the waveform will be t he same, it's just that the frequency is halfed. So that means the period of the graph is doubled.

 

[Shay10825]

What does the change in wavelength do to the graph.

 

[Shay10825]

Well, the waveform will be t he same, it's just that the frequency is halfed. So that means the period of the graph is doubled.

If the period is doubled then why does the graph look the same as the original?

 

[christinono]

Hi everyone .

If I had a sine wave on a string at one instant of time how come when I reduce the wavelength and frequency by half the graph of the wave does not change.

~Thanks

What about the speed of the string?
You must look at the formula v=frequency*wavelength

 

[Curious3141]

Hi everyone .

If I had a sine wave on a string at one instant of time how come when I reduce the wavelength and frequency by half the graph of the wave does not change.

~Thanks

Are you certain that a change in the wavelength will not change the graph of the wave at an instant in time ?

I agree with you about the frequency, though. The instantaneous shape of the graph will not alter with frequency changes.

Do you know the one dimensional wave equation ?

a = a_0\cos(kx - \omega t)

k is the wave number given by \frac{2\pi}{\lambda}

It describes the number of peaks of the wave over a given distance at an instant in time.

\omega is the angular frequency given by 2\pi\nu where \nu is the frequency, or the reciprocal of the period.

It describes the rate at which you will see peaks traveling through a fixed point in space.

Can you see the whole picture now ?

 

[Shay10825]

What about the speed of the string?
You must look at the formula v=frequency*wavelength

The velocity will be multiplied by .25. So why does the graph stay the same??

 

[christinono]

Looking at the equation: v=frequency*wavelength, you can see that the frequency is inversely proportional to the wavelength. Therefore, if you half the frequency, the wavelength is doubled (if the velocity stays constant, of course). Now if you half the wavelength, the frequency doubles. Can you see how these 2 cancel each other out?

 

[christinono]

No, I don't believe it should. If you were simply reducing the wavelenth in half, the graph would look "squished" horizontally by a factor of two. If you were simply reducing the frequency by a factor of 2, the graph would stretch horizontally by a factor of 2 (if all the other variables such as velocity and amplitude were kept constant, of course). But you are doing both, which means the effects cancel each other out.

 

[Shay10825]

Looking at the equation: v=frequency*wavelength, you can see that the frequency is inversely proportional to the wavelength. Therefore, if you half the frequency, the wavelength is doubled (if the velocity stays constant, of course). Now if you half the wavelength, the frequency doubles. Can you see how these 2 cancel each other out?

oh. I see it. Thanks

 

[Shay10825]

a = a_0\cos(kx - \omega t)

What does a and a_0 stand for?

 

[christinono]

i think a is amplitude.

 

[Curious3141]

What does a and a_0 stand for?

a = Displacement (Amplitude) of oscillation at a given point in time and space. It is the vertical coordinate of your graph.

a_0 = Maximum amplitude of oscillation. It is the maximum vertical point your graph reaches.

 

[learningphysics]

The graph should be compressed by a factor of 2. The answer given is wrong. The frequency change does not affect the shape of the graph.

The equation of the wave at a particular time t1 is (modifying Curious' equation):
a = a_0\cos[2\pi(\frac{x}{\lambda} - ft_1)]

a_0 is just the amplitude of the wave.

if you factor out the lambda, it is:

a = a_0\cos[\frac{2\pi}{\lambda}(x - \lambda ft_1)]

So when \lambda becomes 1/2 of the previous value, the coefficient becomes 2, so the graph compresses by a factor of 2.

If f becomes 1/2 of the previous value then the overall shape is unchanged, only the shift along the x-axis is changed.

 

[christinono]

The graph should be compressed by a factor of 2. The answer given is wrong. The frequency change does not affect the shape of the graph.

The equation of the wave at a particular time t1 is (modifying Curious' equation):
a = a_0\cos2\pi(\frac{x}{\lambda} - ft_1)

a_0 is just the amplitude of the wave.

if you factor out the lambda, it is:

a = a_0\frac{\cos2\pi}{\lambda}(x - \lambda ft_1)

So when \lambda becomes 1/2 of the previous value, the coefficient becomes 2, so the graph compresses by a factor of 2.

If f becomes 1/2 of the previous value then the overall shape is unchanged, only the shift along the x-axis is changed.

I don't agree. The wavelength is dependant on the frequency if the speed of the string is kept constant.

 

[learningphysics]

I don't agree. The wavelength is dependant on the frequency if the speed of the string is kept constant.

The question didn't mention speed being constant. Besides, if speed is constant, then it is impossible for both frequency and wavelength to be 1/2 of before.

 

[Shay10825]

I have 1 more question.

For a sinusodial wave on a string, if y= 2sin(4x-3t) then a is?

The answer is:
-18sin (4x-3t)

I don't know where to even start

 

[learningphysics]

I have 1 more question.

For a sinusodial wave on a string, if y= 2sin(4x-3t) then a is?

The answer is:
-18sin (4x-3t)

I don't know where to even start

velocity v=\frac{dy}{dt}

acceleration a=\frac{dv}{dt}=\frac{d^2y}{dt^2}

 

[Shay10825]

ok i understand. Thanks