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.

Question

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., $1080 \mathrm{~km}$, $2468 \mathrm{~km}$, $2160 \mathrm{~km}$, $4320 \mathrm{~km}$

MARKS App · Accepted Answer

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, $d=v_S t_S=4.5 \times 480=2160 \mathrm{~km}$

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