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Phy 232
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** Question Form_labelMessages **
20140713_Q_re_bead_problem
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I think that I've finally figured out how to solve the beads on a string problem, e.g., practice test Problem Number 2: A string is under a tension of 11 Newtons and lies along the x axis. Beads with mass 8.1 grams are located at a spacing of 11 cm along a light but strong string. At a certain instant a certain bead is at y position .002 meters, while the bead to its right is at y position .003 meters and the bead to its left at y position .002 meters. At this instant the bead is moving in the y direction at .0169 m/s. Find the acceleration of the given bead and approximate its velocity .038 seconds later and the distance it will move in this time.
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What I *think* that I understand about the situation:
- Order of solving problem:
- - 1) Calculate net force on bead.
- - 2) Calculate acceleration of bead; result should be same whether one uses f=ma or instead f=-k(displacement), angular velocity = sqrt(k/m), and angular acceleration = -((angular velocity)^2 * displacement).
- - 3) Construct wave equation (setting time and bead in question's x-position to 0 and omitting -kx term as irrelevant, hence position:Y = amplitude * cos([angular velocity * time] - phase shift):
- - -- 3a) Calculate angular velocity (if not done before).
- - -- 3b) Calculate phase shift (phi): (angular acceleration * Position:Y)/Velocity:Y = -tan(phi), so phi = arctan(-(angular acceleration * Position:Y)/Velocity:Y)
- - -- 3c) Calculate amplitude: Initial position = amplitude * cos(-phi), so amplitude = initial position / cos(-phi).
- - 4) Use completed equation to calculate position at specified future time value.
- - 5) Use first derivative of completed equation, i.e., velocity = amplitude * angular velocity * -sin ([angular velocity * time] - phase shift), to calculate velocity at specified future time value.
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1) Does the above procedure look right?
2) Does the solution below look right for the problem as phrased (tension = 11 N; mass of beads = .0085 kg; x-positions of beads = {-11cm, 0, +11cm}; y-positions of beads = {.002 m, .002 m, .003 m}?
- Acceleration of bead:
- - Net force:y on bead = -T sin Theta:1 of string segment vs. x-axis + -T sin Theta:2 of string segment vs. x-axis = -T (sin Theta:1 + sin Theta:2)
- - - Rise:1/Run:1 = 0, so net force:y on bead = -T(sin(arctan[0.002 m / 0.11 m]) = approx. 1.041 deg)) = -(11 N * .0182) approx.= -0.200 N.
- - Calculating acceleration two different ways for confirmation:
- - - Linear calculation: F = ma, so a = F/m = -.200 N / .0081 kg = 24.691 ms^-2.
- - - Calculation using angular velocity:
- - - - Spring constant equivalent (determining restoring force resulting from y-component of string tension): F = -k * displacement; -.200 N = -k * 0.002 m; k = 100.
- - - - Angular velocity = sqrt(k/m) = sqrt(12,345.679 s^-2) = 111.1repeating rad sec^-1.
- - - - So acceleration = -([angular velocity]^2) * distance from equilibrium = -(111.1... rad sec^-1)^2 * 0.002 m = 12345.679 rad sec^-2 * .002 m = -24.691 ms^-2.
- - - These calculations appear to confirm each other.
- Velocity of bead and change of bead's position over interval:
- - Probably-inaccurately-oversimplified calculation using linear approximation assuming acceleration to be constant:
- - - [Never mind; it doesn't work.]
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Actually, this approach will work, and the approach you used below will not (or, rather, it will for only the special cases mentioned in my note below).************
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If you calculate the horizontal and vertical components of the forces pulling each bead to the right and left, as you have done, you can get the net force on each, and dividing by bead mass you can get the bead's acceleration.
At that point you can use a sufficiently small time interval to calculate its change in velocity. Adding this to its initial velocity you get a good approximation to its final velocity (again assuming the interval is small enough), and with the two velocities and the time interval you can also calculate its new position.
Given the state of a string of beads, you can do this for each bead, basing each calculation on the initial position and velocity of each. This will give you the positions and velocities of all the beads after the time interval.
The procedure can then be iterated.
For a bead whose mass is on the order of 10^-2 kg, experiencing forces on the order of .2 N, the magnitude of the acceleration is on the order of 20 m/s^2 and the magnitude of the velocity change on the order of .07 m/s. The bead's average velocity on this time interval would give it a position change of about .0016 meters, if my mental calculation is in the right ballpark.
If we wanted a good model of a string of beads we would probably want to use a shorter time increment so that the changes in position and velocity are less drastic.************
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The slopes of the strings correspond to the first derivative of the waveform with respect to position.
The difference in the slopes (more specifically the slope to the right minus the slope to the left) corresponds to the rate of change of the slope, i.e., to the second derivative of the bead displacement (relative to equilibrium) with respect to position.
The difference in these slopes is directly proportional to the net force on the bead, which is directly proportional to the acceleration of the bead. The acceleration of the bead is the second time derivative of its position.
So the second derivative of bead displacement with respect to time is proportional to the second derivative of its displacement from equilibrium with respect to position.
This is the wave equation.
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- - Using wave equation:
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The wave equation is the equation that states that the second derivative of the y position function with respect to time is in constant proportion to the second dervative of the y position function with respect to space. (see also my notes above)
In a homogeneous medium, ror a standing wave or a traveling wave of constant amplitude and frequency the relatively simply equations we have derived apply.
However the bead situation is not of this nature, so those formulas do not help.
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- - - One needs to get the position equation fully worked out so that one can take its first derivative to get the velocity equation and then its second derivative to get the acceleration equation.
- - - - Position:Y = A * cos (111.1... rad sec^-1 t - kx + phi)...
- - - - - And because the ordinal number of wave that we're on is irrelevant, we can throw out the - kx term...
- - - - - ... yielding Position:y = A * cos (111.1... rad sec^-1 t + phi).
- - - - Next step is to calculate phase shift. Defining time and second bead's x-position as zeroes for those variables:
- - - - - (Omega * Position:Y)/Velocity:Y = -tan(phi).
- - - - - (111.1... rad sec^-1 * .002 m)/(0.0169 ms^-1)...
- - - - - ... so -tan(phi) = 13.149, tan(phi) = -13.149, and phi = -1.495 radians, ...
- - - - - so the equation now is y = A * cos (111.1... rad sec^-1 t - 1.495 rad).
- - - - Next step is to calculate amplitude:
- - - - - At (t = 0, y= 0.002 m), 0.002 m = A * cos(-1.495 rad) and A = 0.002 m / cos(-1.495 rad) = +and-0.0264 m...
- - - - So full equation is Position:Y = 0.0264 m * cos ([111.1... rad sec^-1 * t] - [1.495 rad]).
- - - Velocity:Y = first derivative = -(111.1... * 0.0264 ms^-1) * sin ([111.1... rad sec^-1 * t] - [1.495 rad]).
- - - - I'm not sure whether I'll need it, but just in case: Acceleration:Y = second derivative = -(111.1...^2 * 0.0264 ms^-2) * cos ([111.1... rad sec^-1 * t] - [1.495 rad]).
- - - Velocity at 0.038 seconds = -(111.1... * 0.0264 ms^-1) * sin ([111.1... rad sec^-1 * 0.038 s] - [1.495 rad])...
- - - - ... = -1.181 ms^-1.
- - - Change in y-position = Position:Final - Position:Initial...
- - - - ... = [0.0264 m * cos ([111.1... rad sec^-1 * .038 sec] - [1.495 rad])] - 0.002 m = -0.0242 m - .002 m...
- - - - ... = -0.0262 m.
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Good attempt throughout.
Unfortunately the equations of standing and traveling waves we've used so far don't help, but your original approach does.
Check my notes. Naturally additional questions are welcome.
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