Need help fast with transverse wave on string

[phys62]

A transverse wave is traveling on a string. The displacement y of a particle from its equilibrium position is given by y = (0.0230 m) sin (34.6t - 2.04x). Note that the phase angle 34.6t - 2.04x is in radians, t is in seconds, and x is in meters. The linear density of the string is 1.90 x 10-2 kg/m. What is the tension in the string?

I figured since y=0.0230 sin (34.6t-2.04x) the frequency=34.6 wavelength=3.08 and Amplitude=0.0230
Then I think I can solve for velocity using v=wavelength*frequency and get v=106.568
Now I am stuck as to what I am supposed to do.. I think I have to multiply the amplitude by 4 and then use it somewhere.. Perhaps this equation: v=squareroot of F/(m/L)

But now I am lost! Any help would be greatly appreciated! Thank you!

 

[Thaakisfox]

So your diplacement time-space function is of the form:

$$y(x,t)=A\sin(\omega t -kx)$$

Our wave equation, for this case:

$$\frac{\partial^2 y(x,t)}{\partial t^2}=c^2\frac{\partial^2y(x,t)}{\partial x^2}$$

So the dipersion relation of the simple wave equation is: $$\omega(k)=c k$$
So the phase velocity:

$$c_f=\frac{w(k)}{k}=\frac{\omega}{k}=c$$

So we see the phase velocity of the wave, is equal to the c parameter appearing in the wave equation.

Of this parameter we know that it is (during the derivation of the wave equation for the transverse wave this was what we named c):

$$c=\sqrt{\frac{F}{A \rho}}=\sqrt{\frac{F}{\mu}}$$

Where F is the tension, \rho is the volume mass density, A is the cross sectional area of the string, \mu is the linear mass density.

So we have for the tension:

$$F=\mu c^2=\mu \frac{\omega^2}{k^2}$$

Now put the numerical data in and you are done.

 

[thomate1]

I understand your confusion and I am happy to help. First, let's review the basics of transverse waves. A transverse wave is a type of wave where the disturbance (in this case, the displacement y) is perpendicular to the direction of propagation (in this case, the x-direction). The wave equation for a transverse wave on a string is given by y(x,t) = A sin(kx - ωt), where A is the amplitude, k is the wave number, ω is the angular frequency, x is the position along the string, and t is time.

In this case, we are given the equation y = (0.0230 m) sin (34.6t - 2.04x), which is in the same form as the wave equation. This means that the amplitude A = 0.0230 m, the wave number k = 2.04, and the angular frequency ω = 34.6.

Next, we need to find the velocity of the wave. As you correctly stated, the velocity of a wave is given by v = λf, where λ is the wavelength and f is the frequency. We can find the wavelength by using the wave number, since k = 2π/λ. Therefore, the wavelength is λ = 2π/k = 2π/2.04 = 3.08 m. Now we can find the velocity v = (3.08 m)(34.6 s^-1) = 106.568 m/s.

To find the tension in the string, we can use the equation v = √(T/μ), where T is the tension and μ is the linear density of the string. We are given μ = 1.90 x 10^-2 kg/m, so we can solve for T: T = μv^2 = (1.90 x 10^-2 kg/m)(106.568 m/s)^2 = 204.3 N. Therefore, the tension in the string is 204.3 N.

I hope this helps to clarify the steps you need to take to solve this problem. Remember to always carefully consider the given information and use the appropriate equations to find the unknown values. If you have any further questions, please don't hesitate to ask. Good luck!