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.

Hint:

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.

Answer 1

Correct Answer: 1.19

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: \[ t_1 = \frac{h}{v_p} = \frac{500}{3000} = 0.1667 \, {seconds}. \] where \( v_p = 3000 \, {m/s} \) 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: \[ d = \sqrt{2000^2 + 500^2} = \sqrt{4000000 + 250000} = \sqrt{4250000} \approx 2061.55 \, {m}. \] The travel time for this distance is: \[ t_2 = \frac{d}{v_p} = \frac{2061.55}{3000} \approx 0.6872 \, {seconds}. \] 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: \( t_1 = 0.1667 \, {s} \), 2. The travel time for the offset: \( t_2 = 0.6872 \, {s} \), 3. The wave is reflected twice, so the total time is the sum of the round-trip travel times for both reflections: \[ {Total travel time} = 2 \times (t_1 + t_2) = 2 \times (0.1667 + 0.6872) = 2 \times 0.8539 \approx 1.7078 \, {seconds}. \] Thus, the travel time of the recorded free-surface reflection multiple is approximately: \[ \boxed{1.71 \, {seconds}}. \]