How to work out the reflection point of waves

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SUMMARY

The discussion focuses on calculating the reflection point of waves generated when a workman strikes a steel pipeline. The longitudinal wave speed (vL) is 6.2 km/s, and the transverse wave speed (vT) is 3.2 km/s, with a time difference of 2.4 seconds between the reflected waves. The correct formula derived for the distance to the reflection point is x = 7.9 km, achieved through the equation 2.4 = (2x/Vt) - (2x/Vl), leading to the final calculation of x = 7.936 km.

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  • Understanding of wave mechanics, specifically longitudinal and transverse waves.
  • Familiarity with the speed of sound in different materials, particularly steel.
  • Basic algebra skills for manipulating equations.
  • Knowledge of time-distance-speed relationships in physics.
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  • Learn about the principles of reflection and refraction of waves.
  • Explore advanced topics in wave mechanics, such as wave interference and superposition.
  • Review problem-solving techniques in physics, particularly for wave-related questions.
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Students studying physics, particularly those focusing on wave mechanics, as well as educators seeking to explain wave behavior in materials like steel.

CalinDeZwart
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Homework Statement



When a workman strikes a steel pipeline with a hammer, he generates both longitudinal and transverse waves. The two types of reflected waves return 2.4 s apart. How far away is the reflection point? (For steel, vL = 6.2 km/s, vT = 3.2 km/s).

Homework Equations



Unknown, this is where I need some guidance.

I have referred to the following:
Cutnell, J. D., & Johnson, K. W. (2015). Physics. (10th ed.). New York: John Wiley.
Serway, R.A., Jewett, J.W., Wilson, K., and Wilson, A. (2013). Physics. (Volume 2) (1st ed. Asia‐Pacific
Edition). South Melbourne, Australia: Cengage Learning Australia Pty. Ltd.

3. The Attempt at a Solution

I know the answer is 7.9 km.

Working backwards, I understand it would take V(L) 2.54 seconds and V(T) 4.94 seconds to complete their respective cycles, resulting in the 2.4 second gap.

What I am hoping for is someone who can get me on the right track with a formula to work with.

Thanks
 
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You should be able to arrive at the formula yourself through reasoning. Calling the distance to the reflection time x, how long does it take each wave to go back and forth? What is the difference between those times?
 
Thanks for your reply,

Unfortunately I cannot get my head around how to tackle this question. I know it is basic math, but it just won't click.
 
The best I could come up with t = (d*2) / v
d = (t/2) * v
However, the question doesn't provide data for t and d is the answer I need to arrive at (7.9)

What am I missing?
 
CalinDeZwart said:
The best I could come up with t = (d*2) / v
d = (t/2) * v
However, the question doesn't provide data for t and d is the answer I need to arrive at (7.9)

What am I missing?
Reread my first post and follow the steps it describes. Your equation for the time is correct, but the time is different for each wave - with each traveling at its own speed.
 
I worked it out I think.

2.4 = (2x/Vt) - (2x/Vl)
2.4 = (2x/3.2) - (2x/6.2)
1.2 = (x/3.2) - (x/6.2)

(1.2)(3.2)(6.2) = (6.2x) - (3.2x)

23.808 = 3.0x
x = 23.808/3.0
x = 7.936km
x = [7.9km]

Thanks for your help.