Frequency and wavelength of a wave on a vertical rope

  • Thread starter Helloworld
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Homework Statement


A long, heavy rope hangs straight down from a high balcony on an apartment building. The lower end of the rope hangs about 1.0 m above the ground. If you grab onto the lower end and waggle it back and forth with constant frequency f, a wave travels up the rope. What would happen to the frequency and wavelength of the wave as it travels up the rope? For each property, state whether it would increase, decrease or remain the same, and explain briefly.

Homework Equations


C=√(T/p), C = fλ
f=ω/2π, λ=2π/k

The Attempt at a Solution


The tension increases as we go up the rope since the force at the top is exerted to counteract the weight force of the remaining rope. So the phase velocity C increases meaning that either the frequency or the wavelength must increase to balance the equation C = fλ. The question is, which one will increase? It is stated in the question that the frequency is constant but will it be constant as it travels up the rope?
 

Answers and Replies

  • #2
Orodruin
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Can you perhaps think of any obviously absurd consequences that would result if the frequency was not constant?
 
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Can you perhaps think of any obviously absurd consequences that would result if the frequency was not constant?
Then maybe the frequency will increase because the period decrease(amplitude of the wave decrease as well)?
 
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Orodruin
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I feel as if you are just guessing rather than thinking it through.

Consider a series of pulses from the bottom to the top sent with a separation of 1 s. How far apart would those pulses be at the top of the rope if frequency increased by a factor of 2?
 
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  • #5
I feel as if you are just guessing rather than thinking it through.
Sorry about this. To be honest, I have no clue after the C. I will do more research to answer the question about the pulses
Consider a series of pulses from the bottom to the top sent with a separation of 1 s. How far apart would those pulses be at the top of the rope if frequency increased by a factor of 2?
So T = 1/f and if frequency doubled, T would be 0.5s thus 0.5 seconds apart
 
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