Velocity of Transverse Waves problem

AI Thread 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 derived from the formula involving tension and mass per unit length. Initially, confusion arose regarding the total mass versus mass per unit length, but the solution was clarified by recognizing that the distance (d) equals the length of the cord (l). After substituting and rearranging the equations, the correct distance was found to be approximately 19.25 meters. The discussion highlights the importance of correctly interpreting the variables in wave motion problems.
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:
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
I was thinking using 2 purple mattress samples, and taping them together, I do want other ideas though, the main guidelines are; Must have a volume LESS than 1600 cubic centimeters, and CAN'T exceed 25 cm in ANY direction. Must be LESS than 1 kg. NO parachutes. NO glue or Tape can touch the egg. MUST be able to take egg out in less than 1 minute. Grade A large eggs will be used.
Back
Top