Amplitude of Wave at 5pi/6 Radians in Cycle, y = Asin(kx-wt)

In summary,At π radians with sin wave x = 3mAt 5π radians/6, x = 2.48 mt = 2.48m/6m/sec = .41 secWhat is y at x,tAt t0 x = t = 0
  • #1
morrobay
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Homework Statement


λ = 3m
T = .5 s
k= 2π/λ = 2.09 radians/m
ω= 2π/T = 12.56 = radians/sec
v = ω/k = 6m/sec
At π radians with sin wave x = 3m
At 5π radians/6, x = 2.48 m
t = 2.48m/6m/sec = .41 sec
What is y at x,t
At t0 x = t = 0
[/B]

Homework Equations


y = Asin(kx-ωt)[/B]

The Attempt at a Solution


y Asin (2.09)(2.48) - (12.56)(.41) = 5.18 - 5.14 = sin .04 = .0399 ?
From inspection of a unit circle/sine wave graph with y = A at 900. π/2 radians = 1
.039 above is not correct
https://www.desmos.com/calculator/kxfekf0kgb[/B]
 
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  • #2
"Amplitude of Wave at 5pi/6 Radians in Cycle"
What does this mean? "In cycle" suggests a point in time, but at what point in the x direction?
Please state the problem exactly as given to you.
 
  • #3
All data is in category 1 above : at t = .41 sec , v= 6m/sec x = 2.48 m also 150/180 (3m) = 2.48 m
The question: What is y ( amplitude of sine wave at 1500) , or 5pi/6 radians
A max = 1
www.desmos.com/calculator/cpb0oammx7

(clikck on 14 sine animation)
 
Last edited:
  • #4
morrobay said:
All data is in category 1 above : at t = .41 sec , v= 6m/sec x = 2.48 m also 150/180 (3m) = 2.48 m
The question: What is y ( amplitude of sine wave at 1500) , or 5pi/6 radians
A max = 1
Then my best guess is that it means kx-ωt=5π/6, but then you do not need to know the wavelength or the period.

In your attempt in post #1 you calculated both an x and a t from one piece of information. This cannot be valid.
 
  • #5
haruspex said:
Then my best guess is that it means kx-ωt=5π/6, but then you do not need to know the wavelength or the period.

In your attempt in post #1 you calculated both an x and a t from one piece of information. This cannot be valid.

Why ? I have v from ω/k above so at 2.48 m, t = .41 sec
y ( amplitude) = Asin(kx-ωt) for 5pi/6 ( x = 2.48) radians in the cycle see animation above.
I have shown x = 2.48 m twice above and t=.41 sec also and I repeated question in post # 2
 
  • #6
morrobay said:
so at 2.48 m, t = .41 sec
The 2.48m you calculated as 5/6 of a wavelength. (Actually, you should have found 5/12 of a wavelength; 5π/6 radians is 5/12 of a cycle.)
So 0.41s is the time it takes to travel 5/6 of a wavelength.
You then found the amplitude at offset x at time t. If you think about it, that should have given you zero.
At (0,0) the amplitude is zero, and the wave moves in the positive x direction at speed v, so at (vt,t) it is also zero.

Let's do your calculation more accurately: x=2.5m (exactly)
kx-ωt=(2π/3)2.5-(2π/0.5)(2.5/6)=0.

The question can be made sensible in three ways (at least):
  • specify an x position and ask for the amplitude at time 5T/12.
  • specify a time and ask for the amplitude at 5λ/12.
  • interpret it as asking or A sin(5(2π)/12).
 
  • #7
Then at x = 80m and t = 3.3 s. y = sin 125.9 = .81 And this is valid even when wave has not arrived at 80m ?"
 
  • #8
morrobay said:
Then at x = 80m and t = 3.3 s. y = sin 125.9 = .81
Since the wavelength is 3m, what is happening at 80m is the same as is happenng at the same instant at 2m. Likewise, since the period is .5s, 3.3s is equivalent to 0.3s.
Since the speed is 6m/s, the wave at 2m after 0.3s is the same as the wave at 2-6x0.3 = 0.2m at time 0.
That is 0.2/3 of a wavelength, so (0.2/3)x2π radians or 24 degrees.
morrobay said:
And this is valid even when wave has not arrived at 80m ?"
The wave equation gives you the amplitude at all times and all distances. It is infinite in length, so there is concept of its arrival.
 
  • #9
morrobay said:

Homework Statement


λ = 3m
T = .5 s
k= 2π/λ = 2.09 radians/m
ω= 2π/T = 12.56 = radians/sec
v = ω/k = 6m/sec
At π radians with sin wave x = 3m
At 5π radians/6, x = 2.48 m
t = 2.48m/6m/sec = .41 sec
What is y at x,t
At t0 x = t = 0
[/B]

Homework Equations


y = Asin(kx-ωt)[/B]

The Attempt at a Solution


y Asin (2.09)(2.48) - (12.56)(.41) = 5.18 - 5.14 = sin .04 = .0399 ?
From inspection of a unit circle/sine wave graph with y = A at 900. π/2 radians = 1
.039 above is not correct
https://www.desmos.com/calculator/kxfekf0kgb[/B]
As post#4 suggested, did you try Asin(kx-wt) with A=1 and kx-wt = 5pi/6?
 

1. What is the amplitude of a wave at 5pi/6 radians in a cycle?

The amplitude of a wave at 5pi/6 radians in a cycle is equal to the maximum displacement of the wave from its equilibrium position. In the given equation, y = Asin(kx-wt), the value of A represents the amplitude of the wave.

2. How is the amplitude of a wave related to its energy?

The amplitude of a wave is directly proportional to its energy. This means that the greater the amplitude of a wave, the more energy it carries.

3. Can the amplitude of a wave at 5pi/6 radians be negative?

Yes, the amplitude of a wave at 5pi/6 radians can be negative if the wave is traveling in the negative direction or if the wave is inverted.

4. How does the amplitude of a wave at 5pi/6 radians affect its wavelength and frequency?

The amplitude of a wave at 5pi/6 radians does not have a direct effect on its wavelength or frequency. However, the amplitude, wavelength, and frequency of a wave are all related through the wave speed equation, v = fλ. This means that if the amplitude increases, the wavelength must decrease in order to maintain a constant wave speed.

5. How can the amplitude of a wave at 5pi/6 radians be measured?

The amplitude of a wave at 5pi/6 radians can be measured by using a ruler or measuring device to determine the maximum displacement of the wave from its equilibrium position. This can also be done by measuring the height of the wave on a graph or using specialized equipment such as an oscilloscope.

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