course phy 232
Problem Number 5 A string is under a tension of 15 Newtons and lies along the x axis.
the acceleration of the given bead
its approximate velocity .02 seconds later and
the distance it will move in the .02 seconds.
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so with this i ccould find the velocity of the wave propagation from the induvidual beads from sqrt(tension/bead mass) and consider te bead mass the mass per unit length. is this what im looking for.
to get the equation we solve around for the omega and wave number then dou ble partial derivative w/respect to time
the first velocity i found refers to horizontal velocity so the distnace traveled would be v'dt
am i thinking about this the right way?
The mass of the bead somehow got omitted. Let's assume 4 grams. Its initial velocity in the y direction is also missing. Let's assume initial y velocity 20 cm/s. The positions of the three beads, and their separation, are also missing. Let's assume that they are 5 cm apart, at y positions .0014 cm, .0009 cm (this is the middle bead) and .0018 cm.
You're doing good thinking on this, and this can be reconciled with the solution to the wave equation. However in this case you analyze the motion of the bead based on the y components of the tensions in the strings on either side of it. This process illuminates the derivation of the wave equation, the reasons why it is so.
The string to the right and the string to the left of the bead have very small slopes with respect to the equilibrium axis (the x axis), and therefore both have tension very nearly equal to that of the entire string.
The y component of each tension is easily calculated from the slope of the corresponding string.
What is the slope of the string to the right of the bead? What therefore is the y component of the tension exerted by this string on the bead?
What is the slope of the string to the left of the bead? What therefore is the y component of the tension exerted by this string on the bead?
The x components of the tension are essentially equal, due to the small slope and the near-uniformity of the tension throughout the string.
So what is the net force on the bead?
What is its acceleration?
What are its change in velocity, displacement and new position for the .02 s interval?
A string is under a tension of 11 Newtons and lies along the x axis. Beads with mass 7.1 grams are located at a spacing of 20 cm along a light but strong string. At a certain instant a certain bead is at y position .0026 meters, while the bead to its right is at y position .0019 meters and the bead to its left at y position .003 meters. At this instant the bead is moving in the y direction at .2633 m/s. Find the acceleration of the given bead and approximate its velocity .029 seconds later and the distance it will move in this time.
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We assumed that the strings weight is so light that it can be considered negligible, so each bead has a mass 7.1 g and is 20 cm apart so the mass density is 0.0355 kg/m.
Using the tension we found the velocity of the wave as 17.6028 m/s. We attempted to set up the wave equation placing our middle bead at x = 0 and that t = 0. By using the equation that we set up, we tried to solve for A and w using the position along the x-axis and the y-axis of our other two beads. This created two equations and two unknowns.
When we solved for our angular frequency we got several possible solutions and the one we used was 138.23 rad/s. We then tried to calculate A and got a negative number.
We aren't sure what we are doing wrong in this setup. Are we going in the right direction? Thanks.
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You don't describe your equation, so it's difficult to tell whether it's correct (or whether you are actually using the wave equation or some other equation you've identified as a wave equation--big difference). However your conception is good.
This question can be answered based on just the statics of the string. The string on each side of the bead exerts a force of 11 N, along the direction of the string. The angle of each string relative to horizontal is small, so the sine of the angle has a magnitude very close to the slope (this is because the hypotenuse and adjacent sides are practically the same length, so slope = rise / run = opposite side / adjacent side is practially the same as opposite side / hypotenuse). A carefully drawn and labeled picture which includes the bead and the two sections of string attached to it should make it easy to find the net force. (A quick mental estimate tells me that the force is around .0026 N, so acceleration would be a bit less than 4 m/s^2; if you get this chances are you're right (since we wouldn't be likely to agree if we weren't both right), but if you don't it doesn't mean you're wrong (it could easily be me)).
Let me know what you get, or if you have additional questions.