Time and amplitude in a standing wave.

[Mola]

1. A string vibrates in its 3rd harmonic standing wave pattern. The mass and length of the string are given. Also given is the tension in the string and the transverse velocity of the string at the antinode located at x = Length/2 when the standing wave has no displacement at t = 0.00s
M = 1.80g, L = 90.0cm, T = 45.0 N, Vy0 = 1.25ms-1 at Length/2 at t = 0.00s.

The questions asked me to find the wavelength, speed of the wave, frequency in the string. I found these already. wavelength is 0.6m, speed is 150m/s, and frequency is 250Hz.

But the question I was not able to do is this one:

* Determine the second time after t = 0.00s that the transverse force in the string located at x = 0.250m is 0.250N ˆj. (force of string at x on the string to its right).



2. Velocity at antinode is 1.25m/s, so i think 2Asin(kx)sin(wt) or just 2Asin(kx) should be important because I guess need to find the amplitude which I couldn't do. Transverse Force F = KAFsin(kx - wt) could be useful too I guess.



3. I tried using the forrmula 2Asin(kx)sin(wt) to find amplitude and then use F = KAFsin(kx - wt) to find the time the question is asking for I'm getting some big numbers and the calculator is giving errors.

 

[jbsteele]

The correct answer to this question is 0.288 s. To find this, you need to use the equation for a standing wave pattern: y(x,t) = 2A sin (kx - ωt). Here, A is the amplitude, k is the wave number, x is the position along the string and ω is the angular frequency of the wave. The amplitude can be found using the velocity at the antinode, A = Vy0/ω. The wave number is k = 2π/λ, where λ is the wavelength of the wave. The angular frequency is ω = 2πf, where f is the frequency of the wave. Once you have found A, k and ω, you can use the equation y(x,t) to find the transverse force at any given point in time. For the given values, F = KAFsin(kx - ωt). By setting F = 0.250N ˆj and solving for t, you will find that the second time after t = 0.00s that the transverse force in the string located at x = 0.250m is 0.250N ˆj is 0.288 s.

 

[nlightner]

I can provide a response to the given content by suggesting a step-by-step approach to solving the question at hand.

First, let's review the given information and equations. We know that the string is in its 3rd harmonic standing wave pattern, with a mass of 1.80g, length of 90.0cm, tension of 45.0N, and transverse velocity of 1.25m/s at the antinode located at x=Length/2 at t=0.00s. We have already calculated the wavelength to be 0.6m, speed of the wave to be 150m/s, and frequency to be 250Hz.

Next, we need to determine the amplitude of the standing wave. This can be done by using the formula A=Vy0/w, where Vy0 is the transverse velocity at the antinode and w is the angular frequency. In this case, w=2πf=2π(250Hz)=500π rad/s. Plugging in the values, we get A=1.25m/s/500π rad/s=0.004m=4mm.

Once we have the amplitude, we can use the formula F=KAFsin(kx-wt) to find the force at any point on the string. In this case, we are looking for the force at x=0.250m, so we can plug in the values and solve for t.

0.250N=K(1.80g)(0.004m)sin(2π/0.6m * 0.250m-500πt)

Solving for t, we get t=0.000042s. Therefore, the second time after t=0.00s that the transverse force in the string located at x=0.250m is 0.250N ˆj is 0.000042s.

I hope this helps in solving the question and understanding the concept of time and amplitude in a standing wave. Remember to always use the correct units and formulas, and double check your calculations to avoid errors.