How to work out the reflection point of waves

AI Thread Summary
The discussion revolves around calculating the reflection point of waves generated by striking a steel pipeline. The longitudinal and transverse waves return 2.4 seconds apart, with their respective speeds being 6.2 km/s and 3.2 km/s. The correct approach involves setting up an equation based on the time difference for the two wave types, leading to the formula 2.4 = (2x/3.2) - (2x/6.2). After solving the equation, the reflection point is determined to be approximately 7.9 km away. The conversation emphasizes the importance of understanding the different travel times for each wave type in solving the problem.
CalinDeZwart
Messages
10
Reaction score
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
 
Physics news on Phys.org
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.
 

Thread 'Errors and significant figures'

I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
View full post »
Thread 'A cylinder connected to a hanging mass'

Thread 'A cylinder connected to a hanging mass'

Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
View full post »

Hot Threads

Recent Insights

Back
Top