Calculating Wave Speed and Wavelength for a Vibrating String

AI Thread Summary
To calculate the wave speed and wavelength of a vibrating string with a mass per length of 500 g/m and a tension of 400 N, the formula v = √(T/(m/L)) is used. The user initially miscalculated the wave speed due to unit errors, as they used grams instead of kilograms in their calculations. The correct approach requires converting the mass to kilograms, ensuring proper unit cancellation. After correcting the units, the wave speed can be accurately determined. Understanding these concepts is crucial for solving similar physics problems effectively.
ddsMom
Messages
11
Reaction score
0

Homework Statement



A long string with a mass/length of 500g/m is placed under a tension of 400N. The string is then vibrated up and down with a period of .425sec.

What is the wave speed?

What is the wavelength of the resulting wave?

I have no idea where to beginI would really like to learn this. It's been 20+ years since I have taken a Physics course.
Thank you!


Homework Equations





The Attempt at a Solution



I have no idea where to begin other than this formula:

v=λ/T

I would really like to learn this. It's been 20+ years since I have taken a Physics course.
Thank you!
 
Physics news on Phys.org
what resource are you using? (textbook, on-line notes, ??)
...(here are some very brief notes : http://www.science.marshall.edu/foltzc/211t9.htm )

generally to get a speed, you square root a (Force term divided by an inertia density).
tweak the density so that the units come out right for speed.

sqrt[T/(m/L)].
 
  • Like
Likes 1 person
I did use v=sqrt[T/(m/L)]

so I got v=sqrt[400N/(500g/m)] = .029

Am I on the right track?
 
Changed my g to kg. Thank you so so much for confirming.
 
ddsMom said:
I did use v=sqrt[T/(m/L)]

so I got v=sqrt[400N/(500g/m)] = .029

Am I on the right track?

No. You have made error in calculating square root.
 
  • Like
Likes 1 person
right track, but off by a UNITS error ... recall that Newton is composed of kilo-gram, not gram. (N=kg m/s²)
Always include the units that any numerical value multiplies; kg should cancel inside the sqrt.
 
  • Like
Likes 1 person
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top