How to work out the reflection point of waves

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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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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.