course Phy 121
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15:00:00 Query 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?
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RESPONSE --> Position at clock time x = Acos(`omega* t) Velocity = -`omega *A*sin(`omega* t) Accel = -`omega * A * cos(`omega* t)
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15:00:03 ** Position at clock time is x = Acos(`omega* t) Velocity = -`omega *A*sin(`omega* t) Accel = -`omega * A * cos(`omega* t) University Physics students should note that velocity and acceleration are the first and second derivatives of the position function. **
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RESPONSE --> ok
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15:00:32 How is the acceleration of the pendulum related to the centripetal acceleration of the point on the reference circle?
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RESPONSE --> a = -`omega A sin(`omega *t) and aCent = v^2/r for SHM
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15:00:36 STUDENT 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). **
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RESPONSE --> ok
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15:01:16 How is the kinetic energy of the pendulum related to its restoring force constant k, the amplitude of its motion, and its position x?
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RESPONSE --> The PE of the pendulum at displacement x is .5 k x^2. .5 m v^2 = .5 k A^2 - .5 k x^2 Solve for v v = +- sqrt( .5 k / m * (A^2 - x^2) )
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15:01:20 ** 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) ) . **
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RESPONSE --> ok
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15:02:07 How can we determine the maximum velocity of a pendulum using a washer and a rigid barrier?
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RESPONSE --> If a pendulum of length L is pulled back distance x, and stopped at the equilibrium point a washer on the pendulum will become a projectile to land at a distance determined by the horizontal velocity of the washer. That velocity is the same as the max velocity of the pendulum.
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15:02:10 GOOD 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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RESPONSE --> ok
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15:02:19 Query Gen Phy Problem 11.14 80 N to compress popgun spring .2 m with .18 kg ball.
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RESPONSE --> n/a
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15:02:22 ** 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 / (.18 kg) ) = +- sqrt( 89 m^2 / s^2) = +-9.4 m/s, approx. The speed of the ball is the magnitude 9.4 m/s of the velocity. **
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RESPONSE --> n/a
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15:02:26 Query Gen Phy Problem 11.24 spring 305 N/m amplitude 28 cm suspended mass .260 kg.
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RESPONSE --> n/a
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15:02:29 ** The angular frequency of the oscillation (the angular velocity of the point on the reference circle) is omega = sqrt(k / m), with k = 305 N/m and m = .260 kg. The equation of motion could be y = A sin(omega * t). We obtain omega = sqrt( 305 N/m / (.260 kg) ) = sqrt( 1170 s^-2) = 34 rad/s, approx.. A is the amplitude 28 cm of motion. So the equation could be y = 28 cm sin(34 rad/s * t). The motion could also be modeled by the function 28 sin (34 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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RESPONSE --> n/a
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15:02:34 Univ. 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?
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RESPONSE --> n/a
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15:02:37 GOOD 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). **
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RESPONSE --> n/a
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15:02:47 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> n/a
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