What is the speed of the string? (wave problem)

  • Thread starter Thread starter bored2death97
  • Start date Start date
  • Tags Tags
    Speed String
AI Thread Summary
The discussion centers on calculating the speed of a wave along a spring when the radius is doubled. The initial assumption that doubling the radius results in doubling the mass is incorrect; instead, the mass increases by a factor of four due to the area increase. The correct relationship for wave speed is derived from the tension and linear mass density, leading to the conclusion that the new speed should be v0/2, as stated in the textbook. The error in the original calculation stemmed from misunderstanding the relationship between radius and mass. Clarification on this point is essential for accurate problem-solving in wave mechanics.
bored2death97
Messages
10
Reaction score
0

Homework Statement



A wave travels along a spring at speed v0. What will be the speed if the string is replaced by one made of the same material and under the same tension but having twice the radius.

Answer in book says (v0/2)

Homework Equations



v=√(T/μ)
speed= square root of (tension divided by mu)

μ=m/L
mu= mass/length

The Attempt at a Solution



With the first equation I filled in the equation for mu. Assuming twice the radius would mean twice the mass, I multiplied mass by 2.

v=√[T/(2m/L)]

v=√(TL/2m)

Then if I were to take the value of 2 out of the square root, I would get v=0.707[√(TL/m)]

Rearrange and get v/0.707

So where did I go wrong??
 
Physics news on Phys.org
bored2death97 said:

Homework Statement



A wave travels along a spring at speed v0. What will be the speed if the string is replaced by one made of the same material and under the same tension but having twice the radius.

Answer in book says (v0/2)

Homework Equations



v=√(T/μ)
speed= square root of (tension divided by mu)

μ=m/L
mu= mass/length

The Attempt at a Solution



With the first equation I filled in the equation for mu. Assuming twice the radius would mean twice the mass, I multiplied mass by 2.

v=√[T/(2m/L)]

v=√(TL/2m)

Then if I were to take the value of 2 out of the square root, I would get v=0.707[√(TL/m)]

Rearrange and get v/0.707

So where did I go wrong??

Twice the radius doesn't mean twice the mass. If you double the radius of a circle by what factor does the area increase?
 
Dick said:
Twice the radius doesn't mean twice the mass. If you double the radius of a circle by what factor does the area increase?

It increases by a factor of 4. Thank you.
 
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