Wave Race 2: Explain the logic behind this formulary?

Click For Summary
The discussion revolves around deriving a formula to calculate the distance to an earthquake based on the arrival times of transverse and longitudinal waves. The transverse wave velocity is given as 8.9 km/s, while the longitudinal wave velocity is 5.1 km/s, with transverse waves arriving 73 seconds earlier. The key formula derived is t = [V_longitudinal * Δt] / (V_transverse - V_longitudinal), which relates the time of the transverse wave to the difference in velocities and the time delay. Participants express confusion about the derivation process and emphasize the importance of understanding the underlying algebra. Overall, the conversation highlights the need for clarity in teaching complex physics concepts.
pugfug90
Messages
118
Reaction score
0
https://www.physicsforums.com/showthread.php?t=160610
Continuing off this problem..
My teacher wrote the solution on the board (and conveniently went to the hospital before our period).
Problem again..

Homework Statement


The velocity of the transverse waves produced by an earthquake is 8.9km/s while that of the longitudinal waves is 5.1km/s. A seismograph records the arrival of the transverse waves 73s before that of the longitudinal waves. How far away was the earthquakes?

The Attempt at a Solution


Doc Al's formula was.. the distance/displacement of transverse velocity is equal to the distance/displacement of longitudinal velocity * time + 73s..
Or..
d=vt=(8.9km/s)* T=(5.1km/s) * (T+73s)
And eventually, the T will get isolated..
================================
On the board..
His formula was
Timetransverse=[VlongitudinalΔt]/(vtransverse-Vlongitudinal)

Δt being 73s, the rest being self explanatory.

And it turns out.. seconds is also 98s! Seems like fiddling of numbers until eventually getting the right answer.. but nah:-p

And how in the world did he derive that formula.. Time of transverse wave is equal to the velocity of the longitudinal wave times the change in time divided by the difference in velocity. :bugeye:
 
Last edited:
Physics news on Phys.org
Both waves travel a distance of d.

This involves a little algebra. I'll start the derivation and let you finish it. (It's good practice!)

Consider the following and then see if you can finish what I start and derive his formula.

Transverse Wave:

d=V_Tt

Longitudinal Wave:

d = V_L(t+ \Delta t)

Since both waves travel the same distances, we can eliminate d from the problem as follows:

V_Tt = d = V_L(t+ \Delta t)
V_Tt = V_L(t+ \Delta t)

Now from here on you should be able to isolate t and solve for it. You should get your teachers formula. Good Luck! If you need any more help, don't hesitate to ask.

G01
 
Ahha..
So Vtransverset=Vlongitudinalt + VlongitudinalΔt

Then..

Vtransverset - Vlongitudinalt = VlongitudinalΔt

Then..

t(Vtransverse - Vlongitudinal) =VlongitudinalΔt

Then..

t=[VlongitudinalΔt]/(Vtransverse - Vlongitudinal)

Which is what teacher got. I thought he found some newer innovative way to derive that.. If he had showed how he derived the formula, it would've helped. Thanks:smile:
 
Nice Job. Its no problem. Glad I could help!
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 7 ·
Replies
7
Views
8K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
8
Views
2K
Replies
1
Views
4K
  • · Replies 10 ·
Replies
10
Views
5K
  • · Replies 1 ·
Replies
1
Views
3K
Replies
9
Views
13K
  • · Replies 5 ·
Replies
5
Views
2K