Calculating Frequency and Wavelength of a Sinusoidal Wave on a String

In summary, the problem involves a sinusoidal wave traveling on a string with a given speed and displacement equation. The frequency and wavelength of the wave can be found by considering the period and using the relationship between wave number and wavelength. The angular frequency of the wave can also be determined by the given equation. To find the tension in the string, the wave equation can be used, with the speed and mass per unit length of the string given.
  • #1
mr_coffee
1,629
1
Hello everyone! I'm having troubles understanding why I'm not getting this right...the problem is:

A sinusoidal wave is traveling on a string with speed 10. cm/s. The displacement of the particles of the string at x = 25 cm is found to vary with time according to the equation y = (5.0 cm) sin[10.0 - (4.0 s-1)t]. The linear density of the string is 3.0 g/cm.


What is the frequency of the wave?
Well wouldn't just be 4.0 s^-1?

Also it says What is the wavelength of the wave?
I t hought it would be:
You know the wave number is: (5.0 cm) and you know the relationship:
Wave Length = 2PI/k, where k is the wave number: so i put:
Wave Length = 2PI/5 = 1.2566 which was wrong any ideas why I'm not getting these right?

THanks! :biggrin:
 
Physics news on Phys.org
  • #2
Find the period T first. After one period, the phase has advanced by [itex]2\pi[/itex]. Then use f=1/T.

The wavenumber is not 5.0 cm, that's the amplitude.
Use [itex]\lambda f=v[/itex] to find the speed of the wave.
 
  • #3
mr_coffee said:
What is the frequency of the wave?
Well wouldn't just be 4.0 s^-1?

4.0 is the angular frequency of the wave, [itex]\omega[/itex], in radians/sec; not the frequency, [itex]f[/itex], in cycles/sec. What do you need to multiply radians/sec by, in order to get cycles/sec?
 
  • #4
Thanks for the responce! I'm still kind of confusd on how you find the period with the given information, I'm very rusty on trig. I understand sin has a period of 2PI, but that doesn't mean that function's period is just 1/2Pi does it?

I alwyas found hte period by first finding the angular frequency w, which is suppose to be the number infront of the t. So in this case, would it be 1/4.0, since it says 4.0s^-1?
or is it suppose to be w = -1/4.0? Then i can find the period by using:
f = w/2*pi
T = 1/f

oo my bad, the wave number should be what is infront of the x, but there is no x in the equation :\
I really don't have anything to look at as a reference other then the internet becuase my book doesn't have chapter 16, yes!

Didn't see your message until now jtbell, well 2PI = 360 degree's, so I'm asumming, i would take 4.0 rad/sec *360/2PI ?
But it does say s^-1, so again does this mean its angular frequency is 1/4?
 
Last edited:
  • #5
mr_coffee said:
I alwyas found hte period by first finding the angular frequency w, which is suppose to be the number infront of the t. So in this case, would it be 1/4.0, since it says 4.0s^-1?

Right. The angular frequency [itex]\omega[/itex] is 4.0/s. And this is related to the frequency by [itex]f=2\pi \omega[/itex].

You can get this simply by considering when the phase advances by [itex]2\pi[/itex]. Consider [itex]\sin(\omega t+\phi)[/itex]. One period passes if [itex]\omega T=2\pi[/itex] or [itex]T=2\pi/\omega=1/f[/itex].

And don't fuss with degrees. Use radians.
 
  • #6
Thanks for the help!
f = w/2*pi i think though!
but i got it right once i flip it around. Now i found all the parts to the question but I'm confused on how I'm suppose to find the tension of this string.
I found the following:
What is the frequency of the wave?
.63662
What is the wavelength of the wave?
15.708
Give the general equation giving the transverse displacement of the particles of the string as a function of position and time.
y(x,t) = (5cm)*sin[ (.3999cm^-1)x -(4s^-1)t]
which was correct. Now they said:
What is the tension in the string?

I don't have the book chapter, and in the notes it shows no equation dealing with tension at all. Any ideas? THanks!
 
  • #7
mr_coffee said:
Thanks for the help!
f = w/2*pi i think though!
Yes, you're right :blushing:

A disturbance in a stretched string satisfies the wave equation (for small amplitudes). The speed is:

[tex]v=\sqrt{\frac{T}{\mu}}[/tex].
Where T is the tension in the string and [itex]\mu[/itex] the mass per unit length.
 
Last edited:
  • #8
Thanks again!
hm..i don't see where I'm messing this one up at:
v = 10cm/sec
T = ?
[tex]\mu= 3.0 g/cm[/tex]
if they want Newtons i converted mu to kg/m
3.0 g/cm (1 kg/1000g) (100cm)/1 m = .3 kg/m;
v = .1 m/sec
v^2/[tex]\mu[/tex] = T;
T = .1^2/.3 = .03333N

but that was wrong, i also tried 333.33N by forgetting to convert the 10cm/sec to .1m/sec, any ideas where i f'ed it up?

Oops I'm a tard, it should be v^2*mu

Yep that fixed it! w00t!
 
Last edited:

What is the wave equation?

The wave equation is a mathematical formula that describes the behavior of waves, such as sound waves, light waves, and water waves. It relates the properties of the wave, such as its frequency and wavelength, to the medium it is traveling through.

Why is the wave equation confusing?

The wave equation can be confusing because it involves complex mathematical concepts and equations that may be difficult to understand for those without a strong background in mathematics or physics. Additionally, there are different variations of the wave equation for different types of waves, which can add to the confusion.

What are the applications of the wave equation?

The wave equation has many practical applications in various fields, such as acoustics, optics, and seismology. It is used to study and predict the behavior of waves in these different contexts, and has also been applied in other areas such as signal processing and image analysis.

How is the wave equation derived?

The wave equation is derived using the principles of calculus and physics, specifically the laws of motion and energy conservation. It is a partial differential equation that describes the relationship between the second derivative of a wave's displacement with respect to time and the second derivative of its displacement with respect to distance.

What are some common misconceptions about the wave equation?

One common misconception about the wave equation is that it only applies to transverse waves, when in fact it can also be used to describe longitudinal waves. Another misconception is that the wave equation only applies to simple, idealized waves, when in reality it can also be used to describe more complex wave phenomena. Additionally, some may think that the wave equation can only be used for linear waves, but it can also be applied to nonlinear waves under certain conditions.

Similar threads

  • Introductory Physics Homework Help
Replies
2
Views
854
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
575
  • Introductory Physics Homework Help
Replies
10
Views
986
  • Introductory Physics Homework Help
Replies
9
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
2K
  • Introductory Physics Homework Help
Replies
15
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
804
Back
Top