Earthquakes generate sound waves inside the earth. In case of the earth, both transverse (S) and longitudinal (P) waves can propagate. Typically, the speed of S waves is about 4.5 km `s^-1` and that of P waves is 8.0 km `s^(-1)`. A seismograph records both P and S waves from an earthquake. this difference helps us to find the distanec of the point of origin of the earthquake. this point is called the epicenter.
Q. if at the location of a seismograph the P waevs arrive 2 minute earlier, the distance of the epicenter from the location of the seismograph is

Options

A. 3541.2 km
B. 1234.3 km
C. 2468.6 km
D. 3702.9 km

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the relationship between distance, speed, and time
The distance traveled by both P waves and S waves can be expressed using the formula:
\[ D = V \times T \]
where \( D \) is the distance, \( V \) is the speed, and \( T \) is the time taken.

### Step 2: Define the variables
Let:
- \( D \) = distance from the seismograph to the epicenter
- \( V_p = 8.0 \, \text{km/s} \) (speed of P waves)
- \( V_s = 4.5 \, \text{km/s} \) (speed of S waves)
- \( T_p \) = time taken by P waves to reach the seismograph
- \( T_s \) = time taken by S waves to reach the seismograph

### Step 3: Set up the equations
Since both waves travel the same distance \( D \):
\[ D = V_p \times T_p \]
\[ D = V_s \times T_s \]

From these equations, we can equate the two expressions for \( D \):
\[ V_p \times T_p = V_s \times T_s \]

### Step 4: Relate the times
We know that the P wave arrives 2 minutes (or 120 seconds) earlier than the S wave:
\[ T_s = T_p + 120 \]

### Step 5: Substitute and solve for \( T_p \)
Substituting \( T_s \) in the distance equation:
\[ V_p \times T_p = V_s \times (T_p + 120) \]
Substituting the values of \( V_p \) and \( V_s \):
\[ 8.0 \times T_p = 4.5 \times (T_p + 120) \]

### Step 6: Expand and rearrange the equation
Expanding the right side:
\[ 8.0 T_p = 4.5 T_p + 540 \]
Now, rearranging gives:
\[ 8.0 T_p - 4.5 T_p = 540 \]
\[ 3.5 T_p = 540 \]

### Step 7: Solve for \( T_p \)
Now, dividing both sides by 3.5:
\[ T_p = \frac{540}{3.5} \approx 154.29 \, \text{seconds} \]

### Step 8: Calculate the distance \( D \)
Now that we have \( T_p \), we can find the distance \( D \):
\[ D = V_p \times T_p = 8.0 \times 154.29 \approx 1234.32 \, \text{km} \]

### Final Answer
The distance of the epicenter from the location of the seismograph is approximately:
\[ D \approx 1234.32 \, \text{km} \]

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