Calculate distance from expolision - Waves

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Homework Help Overview

The problem involves calculating the distance from an underground nuclear test explosion based on the time difference between the reception of primary (P) and secondary (S) seismic waves. The speeds of these waves in the Earth's crust are provided, along with the time interval of 68 seconds between their arrivals.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the relevance of various formulas and attempt to clarify the meaning of the time interval in relation to wave speed and distance. There is confusion regarding the use of frequency and the correct equations of motion.

Discussion Status

Some participants have offered guidance on focusing on the relationship between wave speed and the time difference, suggesting a more straightforward approach to the problem. However, there remains a lack of clarity for some participants regarding the application of these concepts.

Contextual Notes

Participants are navigating through the definitions of terms such as frequency and wavelength, and there is an emphasis on understanding the implications of constant speed for the waves involved. The specific values for the times taken by each wave are not known, but the difference is established.

Jason17
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Homework Statement


A recording station observed that there was an interval of 68s between the reception of P (push or primary) waves and S (shake or secondary waves) from an underground nuclear test explosion.
The speed of P and S waves in the Earths crust are 7800 ms^-1 and 4200 ms^-1, respectively.

Calculate the distance of the test site from the explosion


Homework Equations


Speed = Frequency x wavelength (v=fλ)
Speed of electromagnetic radiation = 3 x 10^8 ms^-1
Speed of sound in air 340 ms^-1
Speed of sound in water and sea water = 1500 ms^-1

The Attempt at a Solution


So I am using this formula:
Speed = Frequency x wavelength (v=fλ)
I have the speed, 7800 & 4200 and its frequency i believe? is 68 seconds

Im trying to find the distance (is this the wavelength?)
Wavelength= Speed/Frequency

The answer is 620 km. Which i cannot get to :\

Help appreciated, ty in advance
 
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You don't really need any of the formulas that you wrote.
You know the speed already.
68s is not frequency but the time difference between the detection of the two waves.

Just use the equation for motion with constant speed for each one of the waves.
 


sorry, i don't understand what your trying to say.

Equation of motion?
S=u.t + 0.5.a.t^2 ?
or
S=(u+v)/2 . t?

I still can't get to 620km :(
 


No, the wave has constant speed.
So it's just
d=u1*t1
and
d=u2*t2

where d is the distance between source and receptor and t1, t2 are the propagation times for the two waves.
You don't know t1 and t2 but you know the difference, t1-t2=68 s.
Can you solve the problem now? You need to find d.
 

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