What is the wavelength of a pulse on a hanging rope with changing tension?

[harambe]

Homework Statement

A uniform rope of length 12cm and mass 6kg hangs vertically from a rigid support.A block of mass 2kg is attached to the free end of the rope.A transverse pulse of wavelength 0.06m is produced at the lower end of the rope.What is the wavelength of the rope ,when it reaches the top of the rope

Homework Equations

The Attempt at a Solution

So It is given that the wavelength changes and the tension also changes through out the rope

T(upper segment)= 8g,T(Lower segment)-=2g

The wave equation

v1/v2=(T1/T2)^1/2 {the mass perunit length cancels for this case}

Now velocity of a wave= wavelength X frequency

The wavelength is changing through out the rope and I don't know about frequency so How should I proceed after this...Does the derivation consider the fact that the wave changes velocity at every point on the rope

 

[haruspex]

I don't know about frequency

If 100 complete waves are sent up will they all take the same time to reach the top? What does that tell you about how the frequency at the bottom compares to the frequency at the top?

 

[harambe]

They will take same time so frequency should remain constant

 

[haruspex]

They will take same time so frequency should remain constant

Right, so can you now determine the wavelength change?

 

[palkia]

The frequency remains same so use the relationship between wavelength and velocity to solve this

 

[harambe]

yeah and I got my answer now

 

[pranjal verma]

If 100 complete waves are sent up will they all take the same time to reach the top? What does that tell you about how the frequency at the bottom compares to the frequency at the top?

The elements in the upper part might vibrate less that the elements at the lower part leading to decrease in the frequency.
I still can't understand how your example leads to the explanation that frequency is constant all over the rope.

 

[haruspex]

The elements in the upper part might vibrate less that the elements at the lower part leading to decrease in the frequency.
I still can't understand how your example leads to the explanation that frequency is constant all over the rope.

The waves might change amplitude, but they will not disappear completely, nor will extra waves appear out of nowhere. If the frequency at the bottom is persistently greater, say, than the frequency at the top then then the number of waves in between must be increasing, without limit.

 

[pranjal verma]

If 100 complete waves are sent up will they all take the same time to reach the top? What does that tell you about how the frequency at the bottom compares to the frequency at the top?

I still can't understand how your example leads to the explanation that frequency is constant all over the rope.
The elements in the upper part might vibrate less that the elements at the lower part leading to decrease in the frequency.

The waves might change amplitude, but they will not disappear completely, nor will extra waves appear out of nowhere. If the frequency at the bottom is persistently greater, say, than the frequency at the top then then the number of waves in between must be increasing, without limit.

Thank you very much sir.

 

[haruspex]

The elements in the upper part might vibrate less that the elements at the lower part leading to decrease in the frequency.

That is quite vague, so I am not sure what your reasoning is.
What do you mean by vibrating less? Less amplitude, yes, but less often, no.
Start with one pulse sent from the bottom. It takes some time t to reach the top.
Now send the next pulse. Doesn't that also take time t to reach the top?

 

[pranjal verma]

I agree that all the pulses will take same time to reach the top but how it relates to constant frequency?

 

[haruspex]

I agree that all the pulses will take same time to reach the top but how it relates to constant frequency?

Now consider pulses emitted at times $t_0, t_0+T, t_0+2T, ...$. They arrive at the top at times $t_0+t, t_0+t+T, t_0+t+2T, ...$.

 

[pranjal verma]

So vibration is in vertical direction (up-down) , I thought it was in horizontal axis (left-right).

 

[haruspex]

So vibration is in vertical direction (up-down) , I thought it was in horizontal axis (left-right).

Makes no difference to the argument.

 

[pranjal verma]

Makes no difference to the argument.

I see that time period of all the pulses will be same so all the pulses will have same frequency(up-down vibrations) so it is about vertical vibrations and not horizontal with which I confused .

 

[haruspex]

I see that time period of all the pulses will be same so all the pulses will have same frequency(up-down vibrations) so it is about vertical vibrations and not horizontal with which I confused .

It does not matter whether they are vertical or horizontal. The pulses must arrive at the same rate at which they start, so the frequency is the same.