If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.
Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
036. `query 36
Question: `qQuery class notes #37
If we know the angular frequency `omega and the amplitude A of motion how do we obtain an equation of motion (i.e., the formula that gives us the position of the pendulum if we know the clock time t)? What are the corresponding velocity and acceleration functions?
Your solution:
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Given Solution:
`a** Position at clock time is x = Acos(`omega* t)
Velocity = -`omega *A*sin(`omega* t)
Accel = -`omega^2 * A * cos(`omega* t)
University Physics students should note that velocity and acceleration are the first and second derivatives of the position function. **
UNIVERSITY PHYSICS STUDENTS: DERIVATIVES OF THE POSITION AND VELOCITY FUNCTIONS
The acceleration should be -omega^2 A cos(omega * t).
The velocity function is the derivative of the position function with respect to clock time. That function is a composite of the function A cos(z) with z = omega * t. The derivative of A cos(z) is - A sin(z) and the derivative of omega * t is just omega, so the derivative of the composite is
v(t) = x ' ( t) = (omega * t) ' * (-A sin(omega * t) ) = - omega A sin(omega t)..
The acceleration function is the derivative of the velocity function so again using the chain rule we obtain
a(t) = - omega^2 A cos(omega t).
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Question: `qHow is the acceleration of the pendulum related to the centripetal acceleration of the point on the reference circle?
Your solution:
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Given Solution:
`aSTUDENT ANSWER: a = -`omega A sin(`omega *t) and
aCent = v^2/r for the circle modeling SHM
INSTRUCTOR AMPLIFICATION:
** The centripetal acceleration of the point on the reference circle, which acts toward the center of the circle, has two components, one in the x direction and one in the y direction. The component of the centripetal acceleration in the direction of the motion of the oscillator is equal to the acceleration of the oscillator.
If the oscillator is at position theta then the centripetal acceleration has direction -theta (back toward the center of the circle, opposite to the position vector). The centripetal acceleration is aCent = v^2 / r; so the x and y components are respectively
ax = aCent * cos(-theta) = v^2 / r * cos(theta) and
ay = aCent * sin(-theta) = -v^2 / r * sin(theta). **
STUDENT COMMENT:
I don’t see where the regular acceleration was included in the explanation.
INSTRUCTOR RESPONSE (SUMMARY OF REFERENCE-CIRCLE PICTURE)
The reference circle picture indicates a radial position vector following the
reference point around the circle of radius A at angular velocity omega. The
reference point has a velocity vector tangent to the circle with magnitude omega
* A, and a centripetal acceleration vector directed toward the center of the
circle with magnitude v^2 / r = (omega * A)^2 / A = omega^2 * A.
In this example we are modeling the motion of an oscillator moving along the x
axis. The x components of the position, velocity and acceleration vectors are A
cos(omega t), -omega A sin(omega t) and -omega^2 cos(omega t).
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Question: How is the kinetic energy of the pendulum related to its restoring force constant k, the amplitude of its motion, and its position x?
Your solution:
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Given Solution:
** The PE of the pendulum at displacement x is .5 k x^2.
By conservation of energy, if nonconservative forces are negligible, we find that the KE of the pendulum at position x is.5 k A^2 - .5 k x^2. This result is obtained from the fact that at max displacement A the KE is zero, and the KE change from displacement A to displacement x is the negative of the PE change between these points.
Thus .5 m v^2 = .5 k A^2 - .5 k x^2. Solving for v we have
v = +- sqrt( .5 k / m * (A^2 - x^2) ) . **
STUDENT COMMENT
Ok. So the KE of a pendulum at position x = [1/2(k * A^2) –
1/2(k * x^2)].
INSTRUCTOR RESPONSE
The formula is correct, and it certainly doesn't hurt to know
it and use it.
However the most reliable (and therefore recommended) thing is to understand
where this comes from in terms of energy conservation. If you understand this,
all you need is the formula PE = 1/2 k x^2, and it's unnecessary to clutter up
the landscape with several additional formulas:
The PE at position x is 1/2 k x^2. So we can find the PE change between any two
positions. In the absence of nonconservative force, the PE change is equal and
opposite to the KE change. The rest is just detail:
The PE at x = A is 1/2 k A^2, and since A is the maximum value of x, this is the
maximum PE.
In moving from position A to position x the PE changes from 1/2 k A^2 to 1/2 k
x^2, a decrease, so KE increases by 1/2(k * A^2) – 1/2(k * x^2).
Since KE at position A is zero, it follows that KE at position x is 1/2(k * A^2)
– 1/2(k * x^2).
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Question: `qHow can we determine the maximum velocity of a pendulum using a washer and a rigid barrier?
Your solution:
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Given Solution:
`aGOOD STUDENT ANSWER: If we pullback a pendulum of length L a distance x (much smaller than L), and stop the motion at the equilibrium point (vertical limit of motion) a washer on the pendulum will become a projectile and project off the pendulum, to land at a distance from which we can determine the horizontal velocity of the washer. That velocity is the same as the max velocity of the pendulum, since the max velocity is that which is at the lowest point in its path.
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Question: `qPrinciples of Physics and General College Physics Problem 11.3. Springs compress 5.0 mm when 68 kg driver gets in; frequency of vibration of 1500-kg car?
Your solution:
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Given Solution:
`aFrom the weight of the driver and the compression of the spring, we determine the spring constant (the 'stiffness' of the spring in N / m):
driver weight of 68 kg * 9.8 m/s^2 = 670 N compresses the spring .05 meters, so since | F | = k | x | we have k = | F | / | x | = 670 N / (.005 m) = 134 000 N / m.
Now from the force constant and the mass of the system we have
omega = sqrt(k / m) = sqrt( (134 000 N/m) / (1570 kg) ) = 9 sqrt( (N/m) / kg) ) = 9 sqrt( (kg / s^2) / kg) = 9 rad / s, approximately, or about 1.5 cycles / second.
STUDENT COMMENT
I got the right answer, but when finding ‘omega, I didn’t
combine the car and the
drivers weight…. but I still got 1.5 cycles / s.
INSTRUCTOR RESPONSE
You (correctly) used the driver's weight to calculate k, and
the driver's weight is critical in calculating k.
However the driver's mass is only about 4% of the total mass, and when taking
the square root of the total mass this makes only about a 2% difference in the
result. A 2% difference in a quantity which comes out to about 1.5 probably
won't change the result (2% of 1.5 being only .03).
Bottom line: The driver's mass doesn't change the total mass of the car by a large proportion, so while omitting it makes a difference, it's not a large difference.
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Question: `qPrinciples of Physics and General College Physics problem 11.30: Pendulum with period 0.80 s on Earth; period on Mars, where acceleration of gravity is 0.37 times as great.
Your solution:
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Given Solution:
`aThe period of a angular frequency harmonic oscillator is sqrt(k / m), and the time required for a cycle, i.e., the period of the cycle, is the time required to complete a cycle of 2 pi radians.
For a pendulum we have k = sqrt( m g / L ), where g is the acceleration of gravity. Thus for a pendulum omega = sqrt(k / m) = sqrt( (m g / L) / m) = sqrt( g / L).
From this we see that for a given length, the frequency of the pendulum is proportional to sqrt(g). The period is inversely proportional to the frequency, so the period is inversely proportional to sqrt(g).
Thus we have
period on Mars / period on Earth = sqrt( gravitational acceleration on Earth / gravitational acceleration on Mars) = sqrt( 1 / .37) = 1.7, approximately. So the period on Mars would be about 1.7 * .80 sec = 1.3 sec, approx.
As an alternative to the reasoning or proportionality, we can actually determine the length of the pendulum, and use this length with the actual acceleration of gravity on Mars.
We have
period = 2 pi rad / angular frequency = 2 pi rad / (sqrt( g / L) ) = 2 pi rad * sqrt(L / g). We know the period and acceleration of gravity on Earth, so we can solve for the length:
Starting with period = 2 pi sqrt(L / g)) we square both sides to get
period^2 = 4 pi^2 L / g. Multiplying both sides by g / (4 pi^2) we get
L = g * period^2 / (4 pi^2) = 9.8 m/s^2 * (0.80 sec)^2 / (4 pi^2) = .15 meters.
The pendulum is .15 meters, or 15 cm, long.
On Mars the acceleration of gravity is about 0.37 * 9.8 m/s^2 = 3.6 m/s^2, approx.. The period of a pendulum on Mars would therefore be
period = 2 pi sqrt(L / g) = 2 pi sqrt(.15 m / (3.6 m/s^2)) = 1.3 seconds, approx.
This agrees with the 1.3 second result from the proportionality argument.
STUDENT COMMENT:
I see I should have compared the force of gravity on the two
planets. Why is the period = 2 pi rad/angular frequency?? I
don’t really get that. The rest is just plugging in, so I got that.
INSTRUCTOR RESPONSE:
The angular frequency is the same as the angular velocity of
the reference point. A cycle corresponds to the 2 pi radians around a complete
circle. To find how long it takes to get around the circle, you divide how far
by how fast:
2 pi rad / (angular frequency) = time to complete the cycle.
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Question: `qQuery gen problem 11.14 80 N to compress popgun spring .2 m with .15 kg ball.
Your solution:
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Given Solution:
`a** The PE of the system will be .5 k A^2, where A = .2 m and k = F / x = 80 N / (.2 m) = 400 N / m.
The KE of the released ball will in the ideal case, which is assumed here, be .5 m v^2 = .5 k A^2. Solving for v we obtain
v = +- sqrt( k A^2 / m ) = +- sqrt( 400 N/m * (.2 m)^2 / (.15 kg) ) = +- sqrt( 106 m^2 / s^2) = +-10.3 m/s, approx.
The speed of the ball is the magnitude 10.3 m/s of the velocity. **
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Question: `qQuery gen phy problem 11.24 spring 305 N/m amplitude 28 cm suspended mass .260 kg.
Your solution:
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Given Solution:
`a**The solution given here is for restoring force constant 210 N/m and mass .250 kg. You should be able to adapt your solution accordingly, and you should understand why the angular frequency will be sqrt(305 * .250 / (210 * .260)) times as great as that given here.The angular frequency of the oscillation (the angular velocity of the point on the reference circle) is
omega = sqrt(k / m),
with k = 210 N/m and m = .250 kg.
The equation of motion could be y = A sin(omega * t).
We obtain omega = sqrt( 210 N/m / (.250 kg) ) = sqrt( 840 s^-2) = 29 rad/s, approx..
A is the amplitude 28 cm of motion.
So the equation could be
y = 28 cm sin(29 rad/s * t).
The motion could also be modeled by the function 28 sin (29 rad/s * t + theta0) for any theta0. The same expression with cosine instead of sine would be equally valid, though for any given situation theta0 will be different for the cosine model than for the sine model. **
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Question: `qUniv. 13.74 (13.62 10th edition). 40 N force stretches spring .25 m. What is mass if period of oscillation 1.00 sec? Amplitude .05 m, position and vel .35 sec after passing equil going downward?
Your solution:
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Given Solution:
`aGOOD PARTIAL STUDENT SOLUTION WITH INSTRUCTOR COMMENT
I am sorry to say I did not get that one--but mostly because I am hurrying through these, and I could not locate in my notes, altough I remember doing extensive work through the T=period problems--let me look at Set 9 for a moment.
I think I found something now. If `omega = 2`pi/T and t = 1 sec, `omega = 2pi rad/s
If I convert to accel, thenI can find the mass by way of F = ma.
a = `omega ^2 * A. I do not know A yet so that is no good.
}If A = x then my pullback of x = .25 m would qualify as A, so
a = (2`pi rad/s) ^2 * .25 m = 9.87 m/s^2
So m = F/a = 40.0N/9.87 m/s^2 = 4.05 kg
THAT IS PART A.
INSTRUCTOR COMMENT:
** Good. But note also that you could have found m = k / omega^2 from omega = sqrt(k/m).
F = -k x so 40 N = k * .25 m and k = 160 N/m.
Thus m = 160 N/m / (2 pi rad/s)^2 = 4 kg approx..
STUDENT SOLUTION TO PART B:For part B If A = .050m and T = 1 sec, then the position can be found using the equation, x = A cos(`omega *t)
INSTRUCTOR COMMENT:
** You could model this situation with negative omega, using x = .05 m * sin(-omega * t). This would have the mass passing thru equilibrium at t = 0 and moving downward at that instant.
Then at t = .35 s you would have x = .05 m * sin( - 2 pi rad/s * .35 s ) = .05 m * sin( -.22 rad) = -.040 m, approx..
Velocity would be dx/dt = - 2 pi rad/s * .05 m * cos(-2 pi rad/s * .35 s) = -.18 m/s, approx..
Alternatively you might use the cosine function with an initial angle theta0 chosen to fulfill the given initial conditions:
x = .05 m * cos(2 pi rad/s * t + theta0), with theta0 chosen so that at t = 0 velocity dx/dt is negative and position is x = 0.
Since cos(pi/2) and cos(3 pi/2) are both zero, theta0 will be either pi/2 or 3 pi/2.
The velocity function will be v = dx/dt = -2 pi rad/s * .05 m sin(2 pi rad/s * t + theta0). At t = 0, theta0 = pi/2 will result in negative v and theta0 = 3 pi/2 in positive v so we conclude that theta0 must be pi/2.
Our function is therefore
x(t) = .05 m * cos(2 pi rad/s * t + pi/2).
This could also be written
x(t) = .05 m * cos( 2 pi rad/s * ( t + 1/4 sec) ), indicating a 'time shift' of -1/4 sec with respect to the function x(t) = .05 m cos(2 pi rad/s * t). **
I did the same thing as the student for part B. I think
that’s a right answer it was never said to be incorrect.
INSTRUCTOR RESPONSE:
Your solution, and that of the student, will describe the
motion of the oscillator, but it will not fulfill the conditions of the problem,
in which the oscillator has a certain velocity at a certain instant.
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Question: `qQuery Add comments on any surprises or insights you experienced as a result of this assignment.
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