An earthquake generates both transverse (S) and longitudinal (P) sound waves in the earth. The speed of S waves is about 4.5 km /s and that of P waves is about 8.0 km/s . A seismograph records P and S waves from an earthquake. The first P wave arrives 4.0 min before the first S wave. The epicenter of the earthquake is located at a distance about
An earthquake generates both transverse (S) and longitudinal (P) sound waves in the earth. The speed of S waves is about 4.5 km /s and that of P waves is about 8.0 km/s . A seismograph records P and S waves from an earthquake. The first P wave arrives 4.0 min before the first S wave. The epicenter of the earthquake is located at a distance about
A
25km
B
250km
C
2500km
D
5000km
Text Solution
AI Generated Solution
The correct Answer is:
To solve the problem, we need to find the distance to the epicenter of the earthquake based on the arrival times of the P and S waves.
### Step-by-Step Solution:
1. **Understanding Wave Speeds**:
- The speed of P waves (longitudinal) is given as \( V_P = 8.0 \, \text{km/s} \).
- The speed of S waves (transverse) is given as \( V_S = 4.5 \, \text{km/s} \).
2. **Setting Up the Relationship**:
- Let \( T_S \) be the time taken by the S wave to reach the seismograph.
- Let \( T_P \) be the time taken by the P wave to reach the seismograph.
- According to the problem, the P wave arrives 4 minutes before the S wave:
\[
T_S - T_P = 4 \, \text{minutes} = 240 \, \text{seconds}
\]
3. **Distance Calculation**:
- The distance to the epicenter (D) can be expressed in terms of the time and speed of the waves:
\[
D = V_P \cdot T_P = V_S \cdot T_S
\]
- Therefore, we can write:
\[
D = 8.0 \, T_P \quad \text{(1)}
\]
\[
D = 4.5 \, T_S \quad \text{(2)}
\]
4. **Expressing \( T_P \) in terms of \( T_S \)**:
- From the time difference equation, we can express \( T_P \) as:
\[
T_P = T_S - 240
\]
5. **Substituting \( T_P \) into Equation (1)**:
- Substitute \( T_P \) into equation (1):
\[
D = 8.0 \, (T_S - 240)
\]
- Expanding this gives:
\[
D = 8.0 \, T_S - 1920 \quad \text{(3)}
\]
6. **Setting Equations (2) and (3) Equal**:
- Since both equations represent the same distance \( D \):
\[
4.5 \, T_S = 8.0 \, T_S - 1920
\]
7. **Solving for \( T_S \)**:
- Rearranging gives:
\[
1920 = 8.0 \, T_S - 4.5 \, T_S
\]
\[
1920 = 3.5 \, T_S
\]
\[
T_S = \frac{1920}{3.5} \approx 548.57 \, \text{seconds}
\]
8. **Calculating Distance \( D \)**:
- Now, substitute \( T_S \) back into equation (2):
\[
D = 4.5 \, T_S = 4.5 \times 548.57 \approx 2464.57 \, \text{km}
\]
- Rounding this gives approximately \( D \approx 2500 \, \text{km} \).
### Final Answer:
The epicenter of the earthquake is located at a distance of approximately **2500 km**.
To solve the problem, we need to find the distance to the epicenter of the earthquake based on the arrival times of the P and S waves.
### Step-by-Step Solution:
1. **Understanding Wave Speeds**:
- The speed of P waves (longitudinal) is given as \( V_P = 8.0 \, \text{km/s} \).
- The speed of S waves (transverse) is given as \( V_S = 4.5 \, \text{km/s} \).
...
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