How to work out the reflection point of waves

In summary, when a hammer strikes a pipeline, the waves it produces return 2.4 seconds apart. The distance between the two waves is 7.9 kilometers.
  • #1
CalinDeZwart
10
0

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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  • #2
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?
 
  • #3
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.
 
  • #4
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?
 
  • #5
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.
 
  • #6
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.
 

1. What is a reflection point?

A reflection point is the location where a wave hits a surface and bounces back in the opposite direction.

2. How do I calculate the reflection point of a wave?

The reflection point of a wave can be calculated using the law of reflection, which states that the angle of incidence (the angle at which the wave hits the surface) is equal to the angle of reflection (the angle at which the wave bounces off the surface). This can be done using basic trigonometry.

3. What factors can affect the reflection point of a wave?

The reflection point of a wave can be affected by several factors, including the angle at which the wave hits the surface, the material and texture of the surface, and the frequency and amplitude of the wave.

4. Can the reflection point of a wave be controlled?

Yes, the reflection point of a wave can be controlled by adjusting the angle of incidence, using different materials for the surface, or by using tools such as mirrors or lenses to manipulate the wave's direction.

5. Why is it important to know the reflection point of a wave?

Knowing the reflection point of a wave is important in many fields, such as optics, acoustics, and seismology. It allows us to predict and control the behavior of waves, and is essential in the design and functioning of various devices and systems.

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