Transverse sinusoidal wave on a string

In summary, the wavefunction of a transverse sinusoidal wave on a string has the form y(x,t) = A*cos(kx +omega*t + phi), where x and y are in meters, t is in seconds, and phi is the phase constant in radians. The wave has a period T = 24.2 ms and travels in the negative x direction with a speed of 28.3 m/s. At t = 0, a particle on the string at x = 0 has a displacement of 1.80 cm and travels downward with a speed of 2.05 m/s.
  • #1
K3nt70
82
0
[SOLVED] transverse sinusoidal wave on a string

Homework Statement


The wavefunction of a transverse sinusoidal wave on a string has the form y(x,t) = A*cos(kx +omega*t + phi), where x and y are in m, t is in s and phi is the phase constant in radians. The wave has a period T = 24.2 ms and travels in the negative x direction with a speed of 28.3 m/s. At t = 0, a particle on the string at x = 0 has a displacement of 1.80 cm and travels downward with a speed of 2.05 m/s. What is the amplitude of the wave?



Homework Equations



y(x,t) = Acos(kx + wt + phi)

y(x,t) = A/(((x - vt)^2) + 1)

The Attempt at a Solution



Despite my attempts using these equations, i have made ZERO headway. I think I am having a problem understanding where all the information in the question fits. I did a quick diagram to show what i THINK is happening (is probably not to scale). I guess i have to be PMed for anyone to see it?
 
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  • #2
What is y in these equations? What is A?
 
  • #3
A is amplitude and according to my course notes, "At time t , displacement at given location x is y(x,t)."
 
  • #4
Hi K3nt70,

You can already find k and w, because they give you the period and wave speed. What do you get for those values?

You also need the formula for phi as a funtion of the initial displacement and velocity in the y-direction (at x=0,t=0).

Once you have k, w, and phi, you can solve for A by using your first equation.
 
  • #5
i solved for w fine, (259.64) but i don't know how to solve for k. I know that k = 2pi/lamda but i have no idea what lamda is in this situation..

edit: v = lamda/T gave me lamda,(0.68486) and then i solved for k using k = 2pi/lamda (9.17441). also, in order to use y(x,t) = Acos(kx + wt + phi) don't i need a value for t, time?
 
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  • #6
The problem gives you a value for the displacement at t=0 and x=0, so all you need now is to find phi and you can solve for A.

To find phi, your textbook should have a formula for phi in terms of the tangent of phi and the initial velocity and displacement at x=0. You have to be careful to use the right signs for the values; what do you get for that?
 
  • #7
i can't find any formula similar to what you described above. all the formulas i found with phi also have amplitude in them.
 
  • #8
You have:

[tex]y = A \cos (k x + \omega t + \phi)
[/tex]

If you take the derivative, you find the speed of the particles in the y direction:

[tex]v_y = - A \omega \sin(k x + \omega t + \phi)[/tex]

If you set x=0, and t=0, then you can plug in the displacement and initial speed they give you. What do you get for that?

Then you can take the ratio of the two equations, and convert the sine/cosine to a tangent. You get a standard equation which is something like:

[tex]
\frac{v_y}{y} = -\omega \tan\phi
[/tex]

where the [itex]v_y[/itex] and [itex]y[/itex] on the left hand side are the values at x=0, t=0.
 
  • #9
ok, so I've worked out phi to be -77.2 I am going to put them into the velocity formula and solve for A...

edit: hmm my amplitude came out to be 4.305E-4 m which is wrong.
 
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  • #10
I don't see how you got phi to be -77.2; what numbers are you using to calculate it?
 
  • #11
[tex]
\frac{2.05 m/s}{0.0018 m} = -259.64 *tan\phi
[/tex]

and solved for phi..
 
  • #12
There are three things: The initial velocity at x=0 is negative; your denominator is not correct (there is one too many zeros); also, it's probably best to do everything in radians (it's easiest to plug into the formulas that way).
 
  • #13
okay, so my new phase constant is 0.413368

so... -28.3 m/s = -A(259.64)Sin(0.413368)

i got -0.27 m for A which is wrong. Also, my calculator is now in radians :s
 
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  • #14
Remember that we have two speeds here; the speed on the left hand side of that equation is not the wave speed in the x-direction; it is the speed of the string in the y-direction, and so the number of the left needs to be -2.05 m/s
 
  • #15
YES! thanks for you're help! This led to an addition two correct questions on my assignment.
 

Related to Transverse sinusoidal wave on a string

1. What is a transverse sinusoidal wave on a string?

A transverse sinusoidal wave on a string is a type of mechanical wave that propagates through a medium, in this case a string, by causing particles in the medium to move perpendicular (transverse) to the direction of the wave's motion. A sinusoidal wave has a repeating pattern of crests and troughs and is described mathematically by a sine or cosine function.

2. How is the amplitude of a transverse sinusoidal wave on a string determined?

The amplitude of a transverse sinusoidal wave on a string is the maximum displacement of particles in the medium from their rest position. It is determined by the energy of the wave and the properties of the medium, such as tension and density. A larger amplitude indicates a more energetic wave and a smaller amplitude indicates a less energetic wave.

3. What is the relationship between wavelength and frequency in a transverse sinusoidal wave on a string?

The wavelength of a transverse sinusoidal wave on a string is the distance between two consecutive crests or troughs. The frequency of the wave is the number of complete cycles that pass a given point in one second. The relationship between wavelength and frequency is inversely proportional, meaning that as the wavelength increases, the frequency decreases, and vice versa.

4. How is the speed of a transverse sinusoidal wave on a string calculated?

The speed of a transverse sinusoidal wave on a string is determined by the tension and density of the string. It is calculated using the equation v = √(T/μ), where v is the speed, T is the tension, and μ is the linear density of the string. This means that a higher tension or lower density will result in a faster wave speed.

5. How does a change in tension or density affect a transverse sinusoidal wave on a string?

A change in tension or density can affect the speed, amplitude, and wavelength of a transverse sinusoidal wave on a string. An increase in tension will result in a higher wave speed and larger amplitude, while a decrease in tension will result in a lower wave speed and smaller amplitude. Similarly, an increase in density will result in a lower wave speed and smaller wavelength, while a decrease in density will result in a higher wave speed and larger wavelength.

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