Two wave pulses on a string approach one another at the time t = 0, as shown in the figure below, except that pulse 2 is inverted so that it is a downward deflection of the string rather than an upward deflection. Each pulse moves with a speed of 1.0 m/s. Make a careful sketch of the resultant wave at the times t = 1.0 s, 2.0 s, 2.5 s, 3.0 s, and 4.0 s, assuming that the superposition principle holds for these waves,
To sketch the resultant wave at different times, we need to consider the superposition of wave pulses. When two waves meet, the principle of superposition tells us that the displacements of the waves add up at each point.
Let's start by sketching the initial wave pulses at t = 0:
```
^
| _ _
| / \_/ \
|_____/ \_____
1 2
```
Here, the upward deflection is pulse 1 and the downward deflection is pulse 2.
At t = 1.0 s, both pulses have traveled 1.0 m. Since they have the same speed, they will meet and overlap completely. Additionally, the inverted pulse will be reversed again, resulting in an upward deflection. The resultant wave will be the sum of these two pulses:
```
^
| _ _
| / \_/ \
|_____/ \_____
1 2
```
At t = 2.0 s, pulse 1 has traveled 2.0 m and pulse 2 has traveled 1.0 m. They are not fully overlapping anymore, but we can still add them to obtain the resultant wave:
```
^
| _ _
| / \_/ \
|_______/ \_____
1 2
```
At t = 2.5 s, pulse 1 has traveled 2.5 m and pulse 2 has traveled 1.5 m. The overlap is reduced further, but we can still add the pulses:
```
^
| _ _
| / \_/ \
|_________/ \_____
1 2
```
At t = 3.0 s, pulse 1 has traveled 3.0 m and pulse 2 has traveled 2.0 m. Again, we can add the pulses:
```
^
| _ _
| / \_/ \
|___________/ \_____
1 2
```
At t = 4.0 s, pulse 1 has traveled 4.0 m and pulse 2 has traveled 3.0 m. Adding the pulses:
```
^
| _ _
| / \_/ \
|______________/ \_____
1 2
```
So, at different times, the resultant wave is the superposition of the two wave pulses. The resultant wave gradually separates as time progresses, until the two pulses are fully separated again.
To sketch the resultant wave at different times, we need to understand how the two wave pulses will interfere with each other. Since pulse 2 is inverted, it means it will have an opposite displacement compared to pulse 1.
Let's use the green color to represent pulse 1 and the red color to represent pulse 2. At time t = 0, both pulses are moving towards each other:
```
_______ _______
\ /
_______/ \_______
```
At t = 1.0 s:
Since both pulses are moving with a speed of 1.0 m/s, they will each move 1.0 meter away from their initial position. Thus, the waves will look like this:
```
_______
\ /
_______/ \_______
```
At t = 2.0 s:
Both pulses would have moved 2.0 meters away from their initial position:
```
_______
\ /
_______/ \_______
```
At t = 2.5 s:
Pulse 1 would have moved 2.5 meters away, while pulse 2 would have moved 2.5 meters towards the right. They will overlap partially:
```
______________
\ /
\_______/
```
At t = 3.0 s:
Pulse 1 would have moved 3.0 meters away, while pulse 2 would have moved 3.0 meters towards the right. They will overlap completely:
```
_______ _______
/ \
______/ \_______
```
At t = 4.0 s:
Pulse 1 would have moved 4.0 meters away, while pulse 2 would have moved 4.0 meters towards the right. They will overlap partially:
```
______________
\ /
\___/
```
Keep in mind that this is a qualitative sketch and does not accurately represent the amplitudes or precise shape of the resulting wave. It only shows the general interference pattern that would occur as the two pulses move towards each other.