Calculating Speed and Max Transverse Velocity of a Transverse Wave

In summary, the student is having difficulty understanding a homework question and concept. They try to solve the problem and find that the speed is incorrect.
  • #1
croyboy
3
0
Please help!

Having a little difficulty with a homework question and concept

Homework Statement


y = 2.28sin(0.0276pi x + 2.42pi t)

Find the amplitude, wavelength, frequency, speed and maximum transverse speed.



The Attempt at a Solution



Definitely correct answers

A = 2.28cm
Wavelength = 72.46 cm
Frequency = 1.21 Hz

Questionable answers

Speed = 87.68 cm/s
Max transverse speed = ?


Reasoning

The speed should be w/k which would give 2.42pi/0.0276pi = 87.68 cm/s

The max transverse velocity though...would that be when the trig function of cos is equal to 1 in this equation u = -wAcos(kx-wt) ?
 
Last edited:
Physics news on Phys.org
  • #2
The speed is correct. Transverse speed is when dy/dt is a max. I think you are pretty much nailing that as well.
 
  • #3
Apologies...still not quite sure...would I use the w and A i know multiplied with cos(kx-wt) to get my answer...so if cos(kx-wt)=1, then max transverse speed would be -wA?
 
  • #4
If your displacement function is a sin wave, it's not possible that your velocity "function" be a constant of 87.68cm/s as you calculated. Your speed would be a function as well.

As you know, velocity is the change of displacement over time (i.e. dy/dt), so by differentiating the function for y, you will get a the velocity function (which in this case is a cosine function). That cosine function will allow you to calculate the speed at any location x and time t.

To calculate the maximum speed, we take a look at the velocity function. If you do your calculus right, the equation should look something like:

v = B cos (Cx + Dt)

There is an infinite possibility for the values of x and t, so we can assume that hthe cosine can return any value. The maximum value however of a cosine function is 1. So what does that tell you about your maximum velocity?
 
  • #5
croyboy said:
Apologies...still not quite sure...would I use the w and A i know multiplied with cos(kx-wt) to get my answer...so if cos(kx-wt)=1, then max transverse speed would be -wA?

Yes. Except I'd call the maximum speed +wA. Why would you say minus?
 
  • #6
Thanks much...makes a lot more sense...appreciate the help
 

1. What is the speed of transverse waves?

The speed of transverse waves is the rate at which the wave moves through a medium. It is determined by the properties of the medium, such as density and elasticity, and is not affected by the amplitude or frequency of the wave.

2. How does the speed of transverse waves compare to the speed of longitudinal waves?

The speed of transverse waves is typically faster than the speed of longitudinal waves in the same medium. This is because transverse waves require the particles of the medium to move perpendicular to the direction of the wave, while longitudinal waves require the particles to move parallel to the direction of the wave.

3. Can the speed of transverse waves change?

Yes, the speed of transverse waves can change when they enter a different medium. The change in speed is due to the change in the properties of the medium, such as density and elasticity. This is known as refraction.

4. How does the frequency of a wave affect its speed?

The frequency of a wave does not affect its speed. The speed of a wave is determined by the properties of the medium it is traveling through, not by its frequency. However, the wavelength of a wave is inversely proportional to its frequency, meaning that as the frequency increases, the wavelength decreases.

5. What is the equation for calculating the speed of transverse waves?

The equation for calculating the speed of transverse waves is v = λf, where v is the speed of the wave, λ is the wavelength, and f is the frequency. This equation is known as the wave equation and can be used to calculate the speed of any type of wave, including transverse waves.

Similar threads

  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
9
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
3K
  • Introductory Physics Homework Help
Replies
10
Views
4K
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
10
Views
831
  • Introductory Physics Homework Help
Replies
8
Views
2K
  • Introductory Physics Homework Help
Replies
6
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
2K
Back
Top