An earthquake generates both transverse \(S\) and longitudinal \(P\) waves in the earth with speeds \(4.5…
An earthquake generates both transverse \(S\) and longitudinal \(P\) waves in the earth with speeds \(4.5 \mathrm{~km} \mathrm{~s}^{-1}\) and \(8.0 \mathrm{~km} \mathrm{~s}^{-1}\), respectively. A seismograph records that the first \(P\)-wave arrives 3.5 minutes earlier than the first S-wave. From the seismograph, the epicentre of the earthquake is located at a distance.
\(1080 \mathrm{~km}\)
\(2468 \mathrm{~km}\)
\(2160 \mathrm{~km}\)
\(4320 \mathrm{~km}\)
Solution
Let the distance of epicenter of earthquake from point of observation be \(d\).
speed of \(S\)-wave, \(v_S=4.5 \mathrm{~km}^{-1} \mathrm{~s}\)
speed of \(P\)-wave, \(v_P=8 \mathrm{~km}^{-1} \mathrm{~s}\)
then, \(d=V_P t_P=V_S t_S\)
or \(8 t_P=4.5 t_S\)
\(t_P=\frac{4.5}{8} t_S \quad \ldots (i)\)
The first \(P\)-wave arrives \(3.5 \mathrm{~min}\) earlier than the first \(S\)-wave.
Hence,
\(\begin{aligned}
t_S-t_P & =3.5 \times 60 \\
t_S-t_P & =210 \quad \ldots (ii)
\end{aligned}\)
From Eq. (i), we get
\(\begin{gathered}
t_s-\frac{45}{8} t_s=210 \\
\frac{8 t_S-4.5 t_S}{8}=210 \\
3.5 t_s=210 \times 8 \\
t_S=\frac{210 \times 8}{3.5}=480 \mathrm{~s}
\end{gathered}\)
Now, \(d=v_S t_S=4.5 \times 480=2160 \mathrm{~km}\)