Waves - I just need to confirm an idea

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In summary, the conversation discusses the speed and time of a transverse wave on a uniform rope hanging from a ceiling. The speed is shown to be a function of the distance from the lower end, and the time it takes for the wave to travel the length of the rope is given by a formula involving the gravitational constant and the length of the rope. The participants also mention using the wave speed on a string definition and the period of a simple pendulum in their solutions.
  • #1
DivGradCurl
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Hello everyone!

I've got a question from the book "Fundamentals of Physics/Halliday, Resnick, Walker" - 6th ed, page 395, #23P.

A uniform rope of mass m and length L hangs from a ceiling.

(a) Show that the speed of a transverse wave on the rope is a function of y, the distance from the lower end, and is given by v = sqr(gy).

(b) Show that the time a transverse wave takes to travel the length of the rope is given by T = 2sqr(L/g)

***

(a) Follows from the wave speed on a string definition - v = sqr([tau]/[mu]), where [tau] = mg, and [mu] = m/y --- I think that's right.

(b) I think that the solution might follow from the period of a simple pendulum --- T = sqr(L/g)... but I'm not quite sure how the 2 comes up... it's just a guess.

I hope you guys can give a hand.

Thanks a lot.
 
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  • #2
For (b), you have v = dy/dt = √(gy),

T = integral dt = integral dy/√(gy)

where the latter integral is taken as y ranges from 0 to L.
 
  • #3


Hi there! It looks like you are trying to solve a problem related to waves and their speed and time of travel. I am not familiar with the specific problem from the book you mentioned, but I can offer some general guidance and confirm your ideas.

For part (a), you are correct in using the wave speed on a string formula, v = sqr([tau]/[mu]), where [tau] is the tension in the rope and [mu] is the linear mass density. In this case, [tau] = mg, where m is the mass of the rope and g is the acceleration due to gravity. The linear mass density [mu] can be expressed as m/L since the rope is uniform. Substituting these values in the formula, we get v = sqr((mg)/(m/L)) = sqr(gL/m) = sqr(gy), where y is the distance from the lower end of the rope.

For part (b), your intuition is correct in using the period of a simple pendulum formula, T = 2π sqr(L/g). The 2 comes from the fact that the wave travels up and down the length of the rope, so we need to multiply by 2 to get the total time it takes to travel the length of the rope. Substituting this value in the formula, we get T = 2sqr(L/g).

I hope this helps and good luck with your problem solving! Let me know if you have any other questions or need further clarification.
 

What is a wave?

A wave is a disturbance or oscillation that travels through space and time, transferring energy from one point to another without transporting matter.

How are waves classified?

Waves can be classified based on their direction of movement, the medium they travel through, and the direction of oscillation. The three main types of waves are mechanical waves, electromagnetic waves, and matter waves.

What is the difference between transverse and longitudinal waves?

Transverse waves are characterized by oscillations perpendicular to the direction of wave propagation, while longitudinal waves have oscillations parallel to the direction of propagation. Sound waves are an example of longitudinal waves, while light waves are transverse waves.

What is the relationship between wavelength, frequency, and wave speed?

Wavelength, frequency, and wave speed are all related by the equation v = λf, where v is wave speed, λ is wavelength, and f is frequency. This means that as the wavelength decreases, the frequency increases, and vice versa.

How do waves interact with their environment?

Waves can reflect, refract, diffract, and interfere with other waves and objects in their path. These interactions can change the direction, speed, and amplitude of the waves. The behavior of waves also depends on the properties of the medium they are traveling through.

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