
Earthquakes generate sound waves inside the earth. Unlike a gas, the earth can feel both transverse (S) and longitudinal (P) sound waves. Typically the speed of the S wave is about $ 4km{{s}^{-1}} $ and that of P wave is $ 8km{{s}^{-1}} $ . A seismograph records P and S waves from an earthquake. The first P wave arrives $ 4\min $ before the first S wave. Let us assume that the waves are travelling in a straight line, at what distance will the earthquake occur?
Answer
509.7k+ views
Hint: The concepts of distance and displacement have been used here. Distance is given as the total path length traversed by a body or a wave during the motion. Time taken to travel is given as the velocity divided by the distance travelled. These concepts may help you to solve this question.
Complete step-by-step answer:
Let $ {{V}_{S}} $ and $ {{V}_{P}} $ be the velocities of S and P waves respectively.
Let $ L $ be the distance between the epicentre and the seismograph.
We know that,
$ \begin{align}
& L={{V}_{S}}{{t}_{S}} \\
& L={{V}_{P}}{{t}_{P}} \\
\end{align} $
In which $ {{t}_{S}} $ and $ {{t}_{P}} $ are the respective times taken by the S and P waves in order to reach the seismograph from the epicentre.
It is already mentioned in the question that,
\[\begin{align}
& {{V}_{S}}=4km{{s}^{-1}} \\
& {{V}_{P}}=8km{{s}^{-1}} \\
\end{align}\]
From the equations of the distance between the epicentre and seismograph,
We can write that,
\[{{V}_{S}}{{t}_{S}}={{V}_{P}}{{t}_{P}}\]
Substituting the values in it will give,
\[4\times {{t}_{S}}=8\times {{t}_{P}}\]
Therefore,
\[{{t}_{S}}=2\times {{t}_{P}}\]
And it is also mentioned in the question that,
\[{{t}_{S}}-{{t}_{P}}=4\min \]
Comparing both the equations will give,
\[2\times {{t}_{P}}-{{t}_{P}}=4\min =240s\]
That means,
\[{{t}_{P}}=240s\]
Thus
\[{{t}_{S}}=480s\]
Therefore from the equation regarding length cited above can be written as,
\[L={{V}_{S}}{{t}_{S}}\]
Substituting the values in the equation,
\[L=4\times 480=1920km\]
Therefore the answer is obtained.
Note: Seismic waves are referred to as the waves of energy as a result of the sudden breaking of rock within the earth or by an explosion. They are the energy that can travel through the earth. These are recorded using a seismograph.
Complete step-by-step answer:
Let $ {{V}_{S}} $ and $ {{V}_{P}} $ be the velocities of S and P waves respectively.
Let $ L $ be the distance between the epicentre and the seismograph.
We know that,
$ \begin{align}
& L={{V}_{S}}{{t}_{S}} \\
& L={{V}_{P}}{{t}_{P}} \\
\end{align} $
In which $ {{t}_{S}} $ and $ {{t}_{P}} $ are the respective times taken by the S and P waves in order to reach the seismograph from the epicentre.
It is already mentioned in the question that,
\[\begin{align}
& {{V}_{S}}=4km{{s}^{-1}} \\
& {{V}_{P}}=8km{{s}^{-1}} \\
\end{align}\]
From the equations of the distance between the epicentre and seismograph,
We can write that,
\[{{V}_{S}}{{t}_{S}}={{V}_{P}}{{t}_{P}}\]
Substituting the values in it will give,
\[4\times {{t}_{S}}=8\times {{t}_{P}}\]
Therefore,
\[{{t}_{S}}=2\times {{t}_{P}}\]
And it is also mentioned in the question that,
\[{{t}_{S}}-{{t}_{P}}=4\min \]
Comparing both the equations will give,
\[2\times {{t}_{P}}-{{t}_{P}}=4\min =240s\]
That means,
\[{{t}_{P}}=240s\]
Thus
\[{{t}_{S}}=480s\]
Therefore from the equation regarding length cited above can be written as,
\[L={{V}_{S}}{{t}_{S}}\]
Substituting the values in the equation,
\[L=4\times 480=1920km\]
Therefore the answer is obtained.

Note: Seismic waves are referred to as the waves of energy as a result of the sudden breaking of rock within the earth or by an explosion. They are the energy that can travel through the earth. These are recorded using a seismograph.
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