Help with displacement and velocity of waves on a string (using trig)

[ninjarawr]

Hello all,

excited to be on the forums. I'm having trouble solving a problem from one of my past exams (prepping for final). Could you please guide me?

The displacement associated with a wave on a string has the functional form y = 0.2 cos (10x - 4t). What is the wave speed?

The answer is 0.4 m/s...but how do I get here? what is really bothering me is the "10x"...I'm used to the format y = Acos(wt), where wA = v.


thanks!

ninja

 

[zachzach]

Use the wave equation since it is a wave.

 

[ninjarawr]

Use the wave equation since it is a wave.

yeah...but what do I do with the "10x"?

 

[zachzach]

There is no 10x you are using the differential wave equation right?
$$\frac{\partial^2\psi}{\partial x^2} = \frac{1}{v^2}\frac{\partial^2\psi}{\partial t^2}$$

 

[rdl]

Hello ninja,

I am happy to help with your problem on displacement and velocity of waves on a string using trigonometry. The equation you have provided, y = 0.2 cos (10x - 4t), represents a transverse wave on a string where y is the displacement at a point x and time t. The amplitude of the wave is 0.2 and the angular frequency is 10. The term "10x" in the equation represents the spatial frequency of the wave, indicating how many cycles occur in a distance of x.

To find the wave speed, we can use the formula v = w/λ, where v is the wave speed, w is the angular frequency, and λ is the wavelength. In this case, the wavelength is the distance between two consecutive peaks or troughs of the wave.

To find the wavelength, we can use the trigonometric identity cos (θ) = cos (2π - θ), which means that the cosine function repeats itself every 2π radians. In our equation, the argument of the cosine function is (10x - 4t), which represents the phase of the wave. We can see that when (10x - 4t) = 0, the equation becomes y = 0.2 cos (0) = 0, which means that the wave is at its peak. This happens every time x increases by a value of 2π/10. Therefore, the wavelength is 2π/10 = π/5.

Now, we can plug in the values for w and λ in the formula v = w/λ to get v = 10/(π/5) = 10*5/π = 50/π = 15.92 m/s. However, the units for the amplitude and wavelength are in meters, so we need to convert the units for the wave speed as well. Therefore, the final answer is 0.2*15.92 = 3.184 m/s.

I hope this explanation helps you understand how to solve this type of problem. Let me know if you have any further questions. Best of luck on your final exam!