Question:

Consider a two-dimensional reflection experiment, where a horizontal boundary between two layers is at a depth of 500 m below the free surface. A source and geophone are placed on the free surface with an offset of 2000 m. The P-wave velocity of the top layer is 3000 m/s. The travel time of a recorded free-surface reflection multiple, which got reflected twice at the free surface, is ________ s.

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When calculating the travel time for a free-surface reflection multiple, ensure that you account for both the downward and upward travel paths, as well as the offset between the source and the receiver.
Updated On: Apr 21, 2025
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Solution and Explanation

Step 1: Understanding the situation.
In this two-dimensional reflection experiment, the source and geophone are placed on the free surface, with the boundary at a depth of 500 m. The recorded wave is a multiple reflection, where the wave is reflected twice at the free surface.

Step 2: Analyzing the travel time.
The total travel time for a wave that is reflected twice at the free surface consists of:
- The time for the wave to travel from the source to the boundary at depth \( h = 500 \, {m} \),
- The time for the wave to travel back to the free surface after reflection,
- The time for the wave to travel from the free surface to the receiver, which has an offset of \( 2000 \, {m} \).
The total travel time for the free-surface reflection multiple is the sum of the round-trip travel times, taking into account the path the wave travels and the P-wave velocity of the top layer.

Step 3: Calculating the one-way travel time to the boundary.
The travel time to the boundary is calculated as: \[ t_1 = \frac{h}{v_p} = \frac{500}{3000} = 0.1667 \, \text{seconds}. \] where \( v_p = 3000 \, \text{m/s} \) is the P-wave velocity.

Step 4: Calculating the travel time for the offset.
The travel time for the wave to travel from the source to the receiver with an offset of 2000 m is calculated using the straight-line distance between the source and the receiver. The distance traveled is the hypotenuse of a right triangle with legs \( 2000 \, {m} \) (offset) and \( 500 \, {m} \) (depth). The distance traveled is: \[ d = \sqrt{2000^2 + 500^2} = \sqrt{4000000 + 250000} = \sqrt{4250000} \approx 2061.55 \, \text{m}. \] The travel time for this distance is: \[ t_2 = \frac{d}{v_p} = \frac{2061.55}{3000} \approx 0.6872 \, \text{seconds}. \]

Step 5: Calculating the total travel time.
The total travel time for the recorded free-surface reflection multiple is the sum of the following:
1. The travel time to the boundary: \( t_1 = 0.1667 \, \text{s} \),
2. The travel time for the offset: \( t_2 = 0.6872 \, \text{s} \),
3. The wave is reflected twice, so the total time is the sum of the round-trip travel times for both reflections: \[ \text{Total travel time} = 2 \times (t_1 + t_2) = 2 \times (0.1667 + 0.6872) = 2 \times 0.8539 \approx 1.7078 \, \text{seconds}. \] Thus, the travel time of the recorded free-surface reflection multiple is approximately: \[ \boxed{1.71 \, \text{seconds}}. \]
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