Velocity of Transverse Waves problem

Click For Summary
SUMMARY

The problem involves calculating the distance between two children sending signals along a cord with a total mass of 0.54 kg and a tension of 37 N. The velocity of the transverse wave on the cord is determined using the formula \( v_t = \sqrt{\frac{F_t}{\mu}} \), where \( F_t \) is the tension force and \( \mu \) is the mass per unit length. After resolving the relationship between distance and length, the final calculated distance is approximately 19.25 meters, expressed with two significant figures as 19 m.

PREREQUISITES
  • Understanding of wave mechanics, specifically transverse waves
  • Familiarity with tension force and its application in wave equations
  • Knowledge of mass per unit length calculations
  • Ability to manipulate algebraic equations to solve for unknowns
NEXT STEPS
  • Study the derivation of the wave velocity formula in different mediums
  • Learn about the effects of tension and mass on wave propagation
  • Explore practical applications of transverse waves in engineering
  • Investigate similar problems involving wave speed and distance calculations
USEFUL FOR

Students studying physics, particularly those focusing on wave mechanics, as well as educators looking for examples of problem-solving in wave equations.

rusty65
Messages
4
Reaction score
0

Homework Statement


Two children are sending signals along a cord of total mass 0.54 kg tied between tin cans with a tension of 37 N. It takes the vibrations in the string 0.53 s to go from one child to the other. How far apart are the children?

Express your answer using two significant figures


Homework Equations


Velocity of transverse wave on a cord = sqrt(F_t/\mu)
F_t = Tension Force
\mu = mass per unit length -> m/l

The Attempt at a Solution


I attempted plugging the given values into the formula for velocity of a transverse wave on a cord, and came up with a distance of 4.387 meters. However, after getting the problem wrong (on masterphysics) I realized that the mass given for the cord is its total mass rather than mass per unit length. Seeing as what I am asked to find is the distance between the children (length of the cord) I don't see any way of solving this problem. Am i simply missing the proper formula? Any help would be greatly appreciated.
 
Physics news on Phys.org
Hello rusty65! :smile:

The mass per unit length, as you've written in your relevant equations is m/l. Putting this into the velocity equation and multiplying by time,

vt = t\sqrt{\frac{lF}{{m}}}

What is "vt" in that above equation? :wink:
 
vt is equal to the distance, but the trouble I am having is that the distance, d, that i am attempting to find is equal to the length of the string, l. So i must either be using the wrong formula, or some key piece of information is escaping me.

This is where I am at right now, using the information given:

d = t√((F_t * l)/m) ---plugged in----> d = 0.53√(37l/0.54)

So I've still got two unknowns, d and l, which, according to the wording of the problem, seem to me to be equal to one another. :confused:
 
Scratch that, I figured it out!

Since d = l, I replaced l with d in the equation.

d = 0.53sqrt(37l/0.54) ---> d = 0.53sqrt(37d/0.54)
d/sqrt(d) = 0.53sqrt(37/0.54) ---> d/sqrt(d) = 4.387
d/sqrt(d) = d^(1/2) ---> sqrt(d) = 4.387
d = (4.387)^2
d = 19.246!

Took me a while to get it through my thick head, but I got it now. And thanks for the help!
 
rusty65 said:
Scratch that, I figured it out!

Since d = l, I replaced l with d in the equation.

d = 0.53sqrt(37l/0.54) ---> d = 0.53sqrt(37d/0.54)
d/sqrt(d) = 0.53sqrt(37/0.54) ---> d/sqrt(d) = 4.387
d/sqrt(d) = d^(1/2) ---> sqrt(d) = 4.387
d = (4.387)^2
d = 19.246!

Took me a while to get it through my thick head, but I got it now. And thanks for the help!

Yep! That is what I was suggesting. Good to see you figured it out :smile:
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 10 ·
Replies
10
Views
1K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 10 ·
Replies
10
Views
5K
Replies
13
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K