Question

Hi I reaaaaalllyyy neeed heeeeeelppp to do this lab can anyone please help me and I suppose to find the distance on 4 and 5 there is a Wave motion stimpulation to use for this lab please go to the website provide and I Think you need to copy the link and paste it to the brouser and it open the stimulation thank you please I need it by tonight thank you. Note. you need to follow the first step to adjust the stimpulation thank you

http://phet.colorado.edu/sims/wave-on-a-string/wave-on-a-string_en.html

1.Set the following settings:

Choose Pulse

Set the Amplitude to 50

Set the Damping factor to zero

Choose Fixed End

Check both the Ruler and Time boxes

Set Pulse Width to 50

2.Observe the wave that you’ve created for about thirty seconds. Describe your observations. (Record its measured amplitude and any other observations about what it looks like.

3.What do you think pulse means?

4.How much distance does one pulse travel from end to end?

5.Use the timer to determine how long it takes the pulse to travel the distance measured in #4.

6.Calculate the wave speed (or velocity) with your answers from #4 and #5.

7.Reset the pulse wave back to its starting position and observe what happens as the pulse collides with a fixed end. Record your observations (note any changes in amplitude & direction).

8.Reset the pulse wave back to its starting position and select “Loose End” instead of Fixed End. Now observe what happens as the pulse wave collides with a free end. Record your observations (note any changes in amplitude & direction).

Answer 1

Q1.

all the settings are done as required by the experiment,

Q2.

amplitude of the wave is 25 cm

the wave gets reflected from the fixed end and the reflected wave is inverted in nature.

Q3.

pulse means a short term excitation that acts for a very small duration and only once.

Q4. end to end it travels a distance of 100 cm

Q5. using the timer provided,

time taken to travel from one end to other end is equal to 2.22 seconds.

Q6. wave speed=distance/time

=100 cm/2.22 seconds

=45.045 cm/s

Q7.as the pulse collides with a fixed end, it gets inverted.

the amplitude remains the same as 25 cm but the pulse now has a negative value as compared to positive value before the pulse met the fixed end.

Q8.after switching to loose end,

the pulse does not get inverted and gets reflected in the same form.

the amplitude remains same as before at 25 cm.