Wave Speed on a String - Tension Force, Intensity, Power, Amplitude, Frequency - Inverse Square Law
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- Опубліковано 25 сер 2024
- This physics video tutorial explains how to calculate the wave speed / velocity on a stretch string given an applied tension and linear density of the wire. It also discusses the wave intensity as a function of source power and distance in addition to its relationship to the inverse square law. It covers topics such as period, frequency, and amplitude of a wave. This video contains plenty of examples and practice problems.
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Here is a list of topics:
1. Transverse and Longitudinal Waves
2. Intensity, Amplitude, and Frequency
3. Sound Waves - Compression vs Expansion / Rarefaction
4. High and Low Pressure Regions of Sound Waves
5. Water Waves and Waves on a String
6. Wave Speed on a Stretched Spring
7. Tension Force and Linear Density - Mass Per Unit Length
8. Energy in Wave Motion
9. Wave Intensity, Power, and Distance
10. Inverse Square Law - Intensity and Distance
11. Crest vs Trough of Sinusoidal Wave
12. Period, Frequency, Wavelength, Amplitude and Wave Velocity Calculations
13. Velocity and Tension Force of Spring
14. Principle of Superposition
15. Constructive vs Destructive Interference
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Thank you so much sir, my physics teacher just comes into the class, talks about his personal life, talks about his wife, and his car, and his past. Then gives out homework. You explained it to me much better than my teacher.
Your teacher is talking about the wave of his life
Velocity = Distance/Time
20:20 Intensity
20:30 power
25:30 change of intensity
29:10 Change of intensity with amplitude
thanks saved me some time
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No comments yet? I have one, it is easier sometimes to calculate the ratio of squares as the ratio of their sides to the power of 2, so a^2/b^2 = (a/b)^2, so 30^2/15^2 = (30/15)^2 = (2)^2 = 4; goes a lot faster than 30^2/15^=900/225=4
can you do more question involving strings and energy
I understand well the lesson i realy enjoy it
Thank you for your tutorials! You're a miracle child, lol
And at 17:13, you calculated the Tension product with cm/s not m/s
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Great video - thank you! Does the inverse square law apply to a string as well? Intensity = power/surface area of sphere implies that I α r^2 would only work for waves radiating from a point source in 3D space (radial waves?); does a string count as well?
yeah i guess
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Very informative thank you
how do you calculate Pmax?
Please we need a video on Progressive wave with the two formulas and how to use them on a question
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Okay one question i have is when we cancel out the surface area and the cross sectional area at 49:18 . I mean the surface area of intensity should be the total surface area of all of the sides of the 3D rectangle and the cross sectional area is just one side though. I don't think they can be cancelled. Can someone help me with this please?
can someone explain the proportionate to symbol
Thanks so much!
I guess yaha par bhaut hai jinko coaching se samaj nai aata , fir yaha aate hai hum
In 20:32 does it have to be the surface area of the sphere? Or is it just an example.
Please reply..
Just an example
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I have a question sir, why is density not converted into m³? I mean it should be kg/m³ , why kg/m? thank you
Linear density is not the same as fluid density and is expressed as mass per unit of distance
You said in a previous video the top was the trough and the bottom was a crest....
14:48
around 13:49 T= 2.5s (given) but solving f=distance/time=10m/2.5=4 & 1/4 = 0.25 so T does not equal .25s its 2.5s?
That tension is said to be horizontal component. Not the resultant tension.
39:10
awsome
Everywhere I look it has frequency by itself (not squared) when talking about change of intensity with amplitude or time averaged power, why is it squared in this example? (roughly around 50:00)
You've got to watch the 3hrs video to see how that particular formula is derived
You mean intensity = 2π^2.density.velocity. f^2.A^2
All this other videos on wave are minors
Can you explain how to calcualate the speed of wave??
V=(Lambda)x(Frequency).
Lambda is the wavelength.
For the intensity problem, I divided 125W/4π(1)^2 and got 98.1747 but I am confused because your answer is 9.95.
You probably put it in wrong, you should get 9.95
I don't understand how to solve these qeuetion.........
Q. Consider ocean waves of amplitude 2m and wavelength 200m . Assuming the gravity to be the only active force , the power available per metre perpendicular to the propagation of the wave is....
Ocean wave, means deep water. In this case cg = sqrt(gL/2pi) = sqrt(9.81*200/2pi)
E = (1/16)rho*g*(2A)^2
E = (1/6)(1000)(9.81)(4)
P = E*cg
P = 173.35 kW/m
13:01
Could you please make a video where you focus more so based on theoretical questions? Or does anyone have any good recommendations ?
For theoretical physics, I don't know if you'd like it but check CRASH COURSE.
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