You are given two transverse waves on a
string: f1(x) and f2(x) moving toward each
other. At an initial time, the wave f1(x) (lying
between 0 and 5 m) moves to the right at
v1 = +1 m/s and f2(x) (lying between 5 and
10 m) is moving to the left at v2 = −1 m/s.
What is the speed of the wave after the two
waves have interacted?
The speed of the wave after the two waves have interacted is 0 m/s.
To analyze the behavior of the waves, we need to consider the superposition principle, which states that when two waves meet, the displacement of the resulting wave at any point is the algebraic sum of the displacements of the individual waves.
Given that f1(x) is moving to the right and f2(x) is moving to the left, we can determine how the waves will interact:
1. The waves approach each other:
- The left end of f1(x) is at x = 0 m, and the right end of f2(x) is at x = 10 m.
2. The waves meet each other at x = 5 m:
- At this point, the displacements of f1(x) and f2(x) will add up to determine the resulting wave.
3. Analyzing the displacements of the waves:
- Since f1(x) is moving to the right at v1 = +1 m/s and f2(x) is moving to the left at v2 = -1 m/s, the displacement of f1(x) will be positive (+) and the displacement of f2(x) will be negative (-).
4. Resulting wave between 0 and 5 m:
- As f1(x) moves towards x = 5 m, it will create a positive displacement.
- As f2(x) moves towards x = 5 m, it will create a negative displacement.
- The resulting wave between 0 and 5 m will have a lower amplitude compared to f1(x) alone.
5. Resulting wave between 5 and 10 m:
- As f1(x) moves away from x = 5 m, it will create a negative displacement.
- As f2(x) moves away from x = 5 m, it will create a positive displacement.
- The resulting wave between 5 and 10 m will have a higher amplitude compared to f2(x) alone.
Overall, the waves will superpose and partially cancel each other out between 0 and 5 m, resulting in a lower amplitude wave. Between 5 and 10 m, the waves will combine and create a higher amplitude wave.
To determine what happens when these two waves meet, we need to understand how they add up or interfere with each other.
When the two waves meet, they superpose or combine to create a resultant wave. The resultant wave is the sum of the displacements caused by each individual wave.
To find the resultant wave, we can consider the displacement caused by each wave at different points on the string.
For f1(x), the wave is moving to the right at v1 = +1 m/s. This means that each point on the wave is moving to the right with a velocity of +1 m/s. At the initial time, the wave f1(x) lies between 0 and 5 m.
For f2(x), the wave is moving to the left at v2 = -1 m/s. This means that each point on the wave is moving to the left with a velocity of -1 m/s. At the initial time, the wave f2(x) lies between 5 and 10 m.
As the waves move towards each other, the displacement caused by f1(x) and f2(x) will change depending on the distance from the initial position.
Let's consider a point x = 2 m on the string. At this point, f1(x) has moved to the right by (2 - 0) m = 2 m and f2(x) has moved to the left by (5 - 2) m = 3 m. Therefore, the displacement caused by f1(x) at x = 2 m is 2 m to the right, and the displacement caused by f2(x) at x = 2 m is 3 m to the left.
To find the resultant displacement at x = 2 m, we subtract the displacement caused by f2(x) from the displacement caused by f1(x):
Resultant displacement at x = 2 m = Displacement caused by f1(x) - Displacement caused by f2(x)
= 2 m to the right - 3 m to the left
= -1 m to the left
Similarly, we can find the resultant displacement at other points on the string by subtracting the displacements caused by f2(x) from the displacements caused by f1(x).
Based on the given velocities and initial positions of the waves, we can determine the displacement caused by f1(x) and f2(x) at each point on the string and find the resultant wave.