#$&* course PHY 121 8/3 11 036. `query 36*********************************************
.............................................
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). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Still not sure what it all means. ------------------------------------------------ Self-critique Rating: 1 ********************************************* Question: `qHow is the acceleration of the pendulum related to the centripetal acceleration of the point on the reference circle? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Acceleration of the pendulum is `omega squared (times A cos `omega *t) Centripetal acceleration is velocity squared (divided by r) My understanding is that the point on the reference circle goes around at a steady velocity. The centripetal acceleration is what holds it at an angle and keeps it going in a circle. The pendulum goes through cycles where it slows down then speeds up then slows down then speeds up. The point on the reference circle is going to the right when the pendulum is going to the right, and it's going to the left when the pendulum is going to the left. They meet at 4 places: The point and pendulum coincide on the right and left sides and when the point is at the top and bottom of the reference circle, the pendulum is at equilibrium. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 2 ********************************************* 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: When the pendulum is at an extreme point, which is from equilibrium to amplitude, it is all potential energy. It pauses at velocity zero. It gradually accelerates, changing the PE to kinetic energy until it reaches the point of equilibrium and starts to slow down as it climbs. KE is turning into PE at this point. It continues this cycles. I think that the restoring force constant k is the amount by which the PE and KE are changing back and forth. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): The formula makes sense. ------------------------------------------------ Self-critique Rating: 2 ********************************************* Question: `qHow can we determine the maximum velocity of a pendulum using a washer and a rigid barrier? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The pendulum has a washer sitting on top of it. When the pendulum hits the barrier, the washer falls to the ground under the influence of gravity. Because you know, vertically, v0=0 m/s, a = 9.8 m/s^2, and the vertical distance it dropped, you can figure out how long it took to drop. From that you can figure the horizontal velocity, which would be how fast the pendulum was moving at the point it was abruptly stopped. This is because the washer was moving at the same velocity as the pendulum and continued to travel at that velocity when the pendulum was stopped. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 3 ********************************************* 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The weight of the drivr is 68 kg * 9.8 m/s^2 = 666.4 N k = F/x k = 666.4 N / .005 m k = 1332800 N/m I don't know what to do next. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I have written it in my notebook and understand how the answer was derived. ------------------------------------------------ Self-critique Rating: 2 ********************************************* 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: T = 2 `pi `sqrt L/g .8 s = 2 `pi `sqrt L/9.8 m/s^2 *square both sides* .64 s^2 = 4 `pi^2 L / 9.8 m/s^2 *multiply by 9.8 m/s^2 on both sides* 6.272 = 4 `pi^2 L *divide both sides by 4 `pi^2* L = .1589 m 9.8 m/s^2 * .37 = 3.626 m/s^2 on Mars T = 2 `pi `sqrt L/g T = 2 `pi `sqrt .1589 m/3.626 m/s^2 T = 1.3797 seconds confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qQuery gen problem 11.14 80 N to compress popgun spring .2 m with .15 kg ball. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `a** The PE of the system will be .5 k A^2, where A = .2 m and k = F / x = 80 N / (.2 m) = 400 N / m. The KE of the released ball will in the ideal case, which is assumed here, be .5 m v^2 = .5 k A^2. Solving for v we obtain v = +- sqrt( k A^2 / m ) = +- sqrt( 400 N/m * (.2 m)^2 / (.15 kg) ) = +- sqrt( 106 m^2 / s^2) = +-10.3 m/s, approx. The speed of the ball is the magnitude 10.3 m/s of the velocity. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qQuery gen phy problem 11.24 spring 305 N/m amplitude 28 cm suspended mass .260 kg. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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). ** STUDENT COMMENT: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qQuery Add comments on any surprises or insights you experienced as a result of this assignment. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!