Need Help Solving Water Tank Experiment

  • Thread starter Jack16
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In summary, the conversation discusses an experiment with water waves passing through two openings in a wooden board. The question is where to stand in order to receive little or no wave action. The equation Yk/D = k x lambda /d is used to find the separation between the first "dark" fringe and the central one, with the given values of lambda, D, and x. It is determined that standing at a distance of 100cm from the central direction will result in little to no wave action due to the waves canceling each other out.
  • #1
Jack16
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I Need Some Help:(...

In a water tank experiment,water waves are generated with their srests 2.5 cmapart and parallel.they pass through two openings 5.0 cm apart in a long wooden board.If the end of the tank is 2.0 m beyond the boards,where would you stand ,relative to the "straight - through " direction,so that you received little or no wave action?

this question seems very easy but i just could not end up solvin it ,i used the equation Yk/D = k x lambda /d ,but i couldn't quiet understand how i would find where it would have no wave reaction

I am waiting for your comments,

thanks already...
 
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  • #2
you want have any wave where the two waves which are coherent meet out of phase and cancel each other out...
so from the given data we can see that lambda=2.5cm, D=200cm, x=5cm So you can find the separation the first "dark" fringe has with the central one.
2.5=5y/200 So y=100cm
Although I have not completely understood the tern "Straight through" I think Iam correct
 
  • #3


Hello there,

I can understand your frustration with this experiment. It can be tricky to solve, but let me break it down for you.

First, let's define some variables:
- Yk: distance from the center of one wave crest to the center of the next crest
- D: distance between the two openings in the wooden board (5.0 cm in this case)
- k: wave number, which is equal to 2π/λ (where λ is the wavelength)
- d: distance from the openings to the end of the tank (2.0 m in this case)

Now, the equation you mentioned (Yk/D = kλ/d) is known as the diffraction equation. This equation tells us that the distance between two wave crests (Yk) is equal to the product of the wave number (k) and the wavelength (λ), divided by the distance between the openings (D) and the distance from the openings to the end of the tank (d). This equation is used to calculate the diffraction pattern of the waves as they pass through the openings.

To find where there will be little to no wave action, we need to look at the diffraction pattern. This pattern will show us where the waves interfere constructively (creating larger waves) and where they interfere destructively (canceling each other out).

In this case, since the two openings are 5.0 cm apart, the diffraction pattern will be symmetrical and the first minimum (destructive interference) will occur at an angle of 90 degrees from the straight-through direction. This means that if you stand at a 90-degree angle from the straight-through direction (either to the left or right), you will receive little to no wave action.

I hope this helps you solve the experiment. If you have any further questions, feel free to ask. Good luck!
 

Related to Need Help Solving Water Tank Experiment

1. What is a water tank experiment?

A water tank experiment is a scientific experiment in which a container, typically a tank or basin, is filled with water and used to study various phenomena related to fluid mechanics, such as flow, pressure, and buoyancy. These experiments are commonly used in engineering and environmental science research.

2. How do I set up a water tank experiment?

To set up a water tank experiment, you will need a tank or basin, water, and any materials or instruments needed for your specific experiment. The tank should be filled with water and any additional components, such as pumps or sensors, should be properly installed. It is important to carefully follow the experimental procedure to ensure accurate and reliable results.

3. What are the benefits of conducting a water tank experiment?

Water tank experiments allow scientists to study fluid mechanics in a controlled and repeatable environment. This can provide valuable insights and data for understanding real-world phenomena and developing solutions to engineering and environmental challenges. Additionally, water tank experiments are often more cost-effective and safer than conducting experiments in natural bodies of water.

4. What factors can affect the results of a water tank experiment?

Several factors can affect the results of a water tank experiment, including the design of the tank, the properties of the water, and the accuracy of instruments used to measure variables. Additionally, external factors such as temperature and air pressure can also impact the results. It is important to carefully control and account for these factors in order to obtain reliable data.

5. How can I analyze the data from a water tank experiment?

The analysis of data from a water tank experiment will depend on the specific variables being measured and the goals of the experiment. However, common methods for analyzing data from water tank experiments include plotting graphs, calculating statistical measures, and using mathematical models to interpret the results. It is important to carefully analyze the data to draw accurate conclusions and implications from the experiment.

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