course PHY 231 This is the rest of the work for PHY 231 assignments 34-36. ϵSq؞϶yemŨLassignment #034
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16:28:57 Query Class Notes #33 Why do we say that a pendulum obeys a linear restoring force law F = - k x for x small compared to pendulum length?
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RESPONSE --> According to this equation, x is proportional to F. This is only for an ideal spring(In reality we would have to use some complex calculus, however we are assuming everything is ideal.)
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16:29:01 ** The vertical component of the tension in the string is equal to the weight m * g of the pendulum. At angle `theta from equilibrium we have T cos(`theta) = m * g so T = m * g / cos(`theta). The horizontal component of the tension is the restoring force. This component is T sin(`theta) = m * g * sin(`theta) / cos(`theta) = m * g * tan(`theta). For small angles tan(`theta) is very close to `theta, assuming `theta to be in radians. Thus the horizontal component is very close to m * g * `theta. The displacement of the pendulum is L * sin(`theta), where L is pendulum length. Since for small angles sin(`theta) is very close to `theta, we have for small displacements x = displacement = L * `theta. Thus for small displacements (which implies small angles) we have to very good approximation: displacement = x = L `theta so that `theta = x / L, and restoring force = m * g * `theta = m * g * x / L = ( m*g/L) * x. **
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16:30:15 What does simple harmonic motion have to do with a linear restoring force of the form F = - k x?
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RESPONSE --> Restoring force is equivlant to (4pi^2 *m) / (T^2)
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16:30:17 ** the essential relationship is F = - k x; doesn't matter if it's a pendulum, a mass on an ideal spring, or any other system where net force is a negative constant multiple of the displacement from equilibrium. F = m * a = m * x'', so F = - k x means that m * x'' = - k x. The only functions whose second derivatives are constant negative multiples of the functions themselves are the sine and cosine functions. We conclude that x = A sin(`omega t) + B cos(`omega t), where `omega = `sqrt(k/m). For appropriate C and `phi, easily found by trigonometric identities, A sin(`omega t) + B cos(`omega t) = C sin(`omega t + `phi), showing that SHM can be modeled by a point moving around a reference circle of appropriate radius C, starting at position `phi when t = 0. **
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16:30:52 For a simple harmonic oscillator of mass m subject to linear net restoring force F = - kx, what is the angular velocity `omega of the point on the reference circle?
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RESPONSE --> The angular velocity omega is equivalent to the square root of force constant k divided by mass m
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16:30:58 STUDENT RESPONSE: omega= sqrt (k/m) INSTRUCTOR COMMENT: Good. Remember that this is a very frequently used relationship for SHM, appearing in one way on another in a wide variety of problems.
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16:31:39 If the angular position of the point on the reference circle is given at clock time t by `theta = `omega * t, then what is the x coordinate of that point at clock time t?
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RESPONSE --> It would be cos 0. cos 0 = 1, which would keep this as is.
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16:31:42 since theta=omega t, if we know t we find that x = radius * cos(theta) or more specifically in terms of t x = radius*cos(omega*t)
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16:32:38 Query introductory problem set 9, #'s 1-11 If we know the restoring force constant, how do we find the work required to displace the oscillator from its equilibrium position to distance x = A from that position? How could we use this work to determine the velocity of the object at its equilibrium position, provided we know its mass?
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RESPONSE --> We know that work is equivalent to worce * displacement. Once we find the magnitude of force, we can solve for work assuming we can find or have the value of the displacement.
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16:32:40 ** You can use the work 1/2 k A^2 and the fact that the force is conservative to conclude that the max PE of the system is 1/2 k A^2. This PE will have transformed completely into KE at the equilibrium point so that at the equilibrium point KE = .5 m v^2 = .5 k A^2. We easily solve for v, obtaining v = `sqrt(k/m) * A. ** STUDENT COMMENT: I'm a little confused by that 1/2 k A^2. INSTRUCTOR RESPONSE: That is the PE at x = A. To directly reason out the expression PE = .5 k A^2 we proceed as follows: PE = work done by system in moving from equilibirum * displacement = fAve * `ds. The force exerted on the system at position x = A is -k A. The force exerted at position x = 0 is 0. Force changes linearly with position. So the average force exerted on the system is ( 0 - kA) / 2 = -1/2 k A. The force exerted by the system is equal and opposite, so fAve = 1/2 k A. The displacement from x = 0 to x = A is `ds = A - 0 = A. We therefore have PE = fAve `ds = 1/2 k A * A = 1/2 k A^2. This is also the area beneath the F vs. x graph between x = 0 and x = A. That region is a triangle with base A and altitude k A so its area is 1/2 base * height = 1/2 k A^2.
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16:33:05 Query Add comments on any surprises or insights you experienced
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RESPONSE --> This was interesting evaluating the formulas from a different perspective.
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16:33:08 as a result of this assignment.
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ʂy assignment #035 骶͌Άrzvc֡ Physics I 10-12-2008
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16:59:18 Query introductory problem set 9, #'s 12-17 How do we find the mass of a simple harmonic oscillator if we know the restoring force constant k and how long it takes to complete a cycle?
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RESPONSE --> We have T and k. T = 2pi * sqroot(m/k) square both sides to get T^2 = 4pi^2 * m/k multiply both sides by k to get T^2 * k = 4pi^2 * m divide by 4pi^2 to get m = (T^2 * k) / (4pi^2)
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16:59:23 We know the basic relationship omega = sqrt(k/m), which we can solve to get m = omega^2 * k. We are given k, so if we know omega we can easily find m. We know how long it takes to complete a cycle so we can find the angular frequency omega: From the time to complete a cycle we find the frequency, which is the reciprocal of the time required. From frequency we find angular frequency omega, using the fact that 1 complete cycle corresponds to 2 pi radians.
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17:02:11 If we know the mass and length of a pendulum how can we find its restoring force constant (assuming displacements x much less than pendulum length)?
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RESPONSE --> We can move k to one side and solve for it using the following equation: k = (4pi^2) / T^2 If we can find T, we can find k. We have m and L, which is more than enough. T = 2pi * sqroot(L/g). We have L to find T, then we substitute the other values.
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17:02:13 For small displacement from equilibrium vector resolution of the forces tells us that the x component of tension in the same proportion to tension as the displacement x to the length L: x / L Since for small angles the tension is very nearly equal to the weight mg of the pendulum this gives us Tx / m g = x / L so that Tx = (m g / L) * x. Since Tx is the restoring force tending to pull the pendulum back toward equilibrium we have restoring force = k * x for k = m g / L. So the restoring force constant is m g / L.
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17:02:32 Query Add comments on any surprises or insights you experienced
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RESPONSE --> I am confident that I understand this material.
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17:02:34 as a result of this assignment.
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ᛔ[g˝MaeWL}}}[ assignment #036 骶͌Άrzvc֡ Physics I 10-12-2008
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17:04:32 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 --> If we need motion, or velocity, then we have an equation. Vmax = A * omega
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17:05:44 ** 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 --> I assumed that the trigonometric values would equal 1. cos 0 =1 sin 90 = 1
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17:07:40 How is the acceleration of the pendulum related to the centripetal acceleration of the point on the reference circle?
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RESPONSE --> centripetal acceleration = v^2 / r Acceleration of a pendulum = -omegaAcos(omega*t) We include omega for a pendulum because it does not have uniform motion. If a pendulum made a complete circle, we could use the centripetal acceleration equation. Again, it does not move uniformly, so we need to use the angular velocity.
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17:08:19 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 --> I said cosine instead of sine.
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17:09:28 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 --> Kinetic Energy KE = 1/2mv^2 F = kx I am actually not sure unless I put in all of the SHM equations and find their relationships.
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17:10:12 ** 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 --> I should have known the PE equation.
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17:11:26 How can we determine the maximum velocity of a pendulum using a washer and a rigid barrier?
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RESPONSE --> Vmax = omega * A We can find its angular velocity by multiplying its frequency by 2pi (after finding the period in the washer). Then we can find the amplitude and simply plug and chug.
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17:11:29 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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17:11:33 Principles of Physics and General College Physics Problem 11.3. Springs compress 5.0 cm when 68 kg driver gets in; frequency of vibration of 1500-kg car?
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17:11:35 From 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 / (.05 m) = 13,400 N / m. Now from the force constant and the mass of the system we have omega = sqrt(k / m) = sqrt( (13,400 N/m) / (1570 kg) ) = 3 sqrt( (N/m) / kg) ) = 3 sqrt( (kg / s^2) / kg) = 3 s^-1, or 3 cycles / second.
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17:11:37 Principles 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.
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17:11:39 The 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.
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17:11:44 Query gen problem 11.14 80 N to compress popgun spring .2 m with .15 kg ball.
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17:11:46 ** 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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17:11:49 Query gen phy problem 11.24 spring 305 N/m amplitude 28 cm suspended mass .260 kg.
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17:11:51 **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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17:26:04 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 --> A) We need the mass that must be suspended from the spring. Let's use the following equation: T = 2pi * sqroot(m/k) We have T, we need m. We first need to find k. F = kx (we have F and x) k = F/x k = 40 N / 0.250m k = 160 N /m Now we need to use some algebra to get m by itself in the original equation. T = 2pi * sqroot(m/k) T^2 = 4pi^2 * m/k T^2 * k = 4pi^2 * m (T^2 * k) / (4pi^2) = m (1sec^2 * 160N/m) / (39.478) = m so m = 4.053kg Part B ) t = (0.35sec)*T x = -Asin2pi(0.35) x = -0.050msin2pi(0.35) x = -0.0405m The motion is upward Part C) Fsping -mg = -kx Fspring = -(160N/m)(-0.030m) + (4.05 kg)(9.8m/sec^2) = 44.5N
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17:26:35 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 --> Am I missing part C
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17:26:51 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> Part B and C were tricky towards the end/
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course PHY 231 This is the rest of the work for PHY 231 assignments 34-36. UԖז_Eassignment #035
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16:41:26 `q001. Note that this assignment contains 5 questions. At its maximum velocity, a simple harmonic oscillator matches the speed of the point moving around its reference circle. What is the maximum velocity of a pendulum 20 cm long whose amplitude of oscillation is 20 cm? Note that the radius of the reference circle is equal to the amplitude of oscillation.
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RESPONSE --> Maximum velocity v is found using the following equation: Vmax = omega * A
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16:41:52 We need to find the velocity of the point on the reference circle that models this motion. The reference circle will have a radius that matches the amplitude of oscillation, in this case 20 cm. The period of the oscillation is T = 2 `pi `sqrt( L / g ) = 2 `pi `sqrt( 20 cm / (980 cm/s^2) ) = .9 sec, approx.. Thus the point completes a revolution around the reference circle once every .9 sec. The circumference of the reference circle is 2 `pi r = 2 `pi * 20 cm = 126 cm, approx., so the point moves at an average speed of 126 cm / .9 sec = 140 cm/s. Thus the maximum velocity of the pendulum must be 140 cm/s.
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RESPONSE --> My answer was in m/sec instead, but that seems to be correct still. self critique assessment:
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16:46:33 `q002. If the 10 kg mass suspended from the 3000 N/m spring undergoes SHM with amplitude 3 cm, what is its maximum velocity?
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RESPONSE --> This problem has the same approach as the last one except that we need to find another way to find period T, which we can do. We have mass m and force constant k. T = 2pi*sqroot(m/k) T = 2pi*sqroot(10kg/3000N/m) T = 2pi*sqroot(0.00333) T = 2pi* 0.0577 T = 0.3627sec Now I will use the same steps that I did before. T = 2pi/omega omega = 2pi / T omega = 2pi / 0.3627sec omega = 17.323 rad/sec Now let's plug this into our original equation: Vmax = omega * A Vmax = 17.323rad/sec * 0.03m Vmax = 0.52 m /sec confidence assessment: 3
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16:47:01 We previously found the angular frequency and then the period of this system, obtaining period of oscillation T = .36 second. The reference circle will have radius 3 cm, so its circumference is 2 `pi * 3 cm = 19 cm, approx.. Traveling 19 cm in .36 sec the speed of the point on the reference circle is approximately 19 cm / (.36 sec) = 55 cm/s. This is the maximum velocity of the oscillator.
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RESPONSE --> Again my answer was in meters /sec but it works out. It feels very good to have the right answer. self critique assessment: 2
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16:48:54 `q003. What is the KE of the oscillator at this speed?
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RESPONSE --> We have all the information we need to solve for KE KE = 1/2mv^2 m was 10kg and v = 0.52m/sec So KE = 1/2(10kg) (0.52m/sec)^2 KE= 5kg * 0.27m^2/sec^2 KE = 1.35 J confidence assessment: 2
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16:49:22 The KE is .5 m v^2 = .5 * 10 kg * (.55 m/s)^2 = 1.5 Joules, approx.. Note that this is the maximum KE of the oscillator.
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RESPONSE --> alright my answer appears to be close enough. My work is correct. self critique assessment: 2
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16:51:14 `q004. How much work is required to displace the mass 3 cm from its equilibrium position?
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RESPONSE --> Work is found multiplying an object's force by its displacement. We do not have force(unless we can use the force constant) But change in kinetic energy is equivalent to work. If this is true, shouldn't the answer be the same as the previous problem? (1.35J) confidence assessment: 2
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16:52:08 The mass rests at its equilibrium position so at that position there is no displacing force, since equilibrium is the position taken in the absence of displacing forces. As it is pulled from its equilibrium position more and more force is required, until at the 3 cm position the force is F = k x = 3000 N / m * .03 m = 90 N. (Note that F here is not the force exerted by the spring, but the force exerted against the spring to stretch it, so we use kx instead of -kx). Thus the displacing force increases from 0 at equilibrium to 90 N at 3 cm from equilibrium, and the average force exerted over the 3 cm displacement is (0 N + 90 N ) / 2 = 45 N. The work done by this force is `dW = F `dx = 45 N * .03 m = 1.5 Joules.
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RESPONSE --> My theory seems to be correct in this case. self critique assessment: 2
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16:54:31 `q005. How does the work required to displace the mass 3 cm from its equilibrium position compare to the maximum KE of the oscillator, which occurs at its equilibrium position? How does this result illustrate the conservation of energy?
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RESPONSE --> Work is equivalent to change in KE. KE final - KE initial = work. Unfortunately, I do not have the value for KE initial to solve for this. This demonstrates conservation of energy because nothing is created nor destroyed. confidence assessment: 1
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16:54:46 Both results were 1.5 Joules. The work required to displace the oscillator to its extreme position is equal to the maximum kinetic energy of the oscillator, which occurs at the equilibrium position. So 1.5 Joules of work must be done against the restoring force to move the oscillator from its equilibrium position to its extreme position. When released, the oscillator returns to its equilibrium position with that 1.5 Joules of energy in the form of kinetic energy. Thus the work done against the restoring force is present at the extreme position in the form of potential energy, which is regained as the mass returns to its equilibrium position. This kinetic energy will then be progressively lost as the oscillator moves to its extreme position on the other side of equilibrium, at which point the system will again have 1.5 Joules of potential energy, and the cycle will continue. At every point between equilibrium and extreme position the total of the KE and the PE will in fact be 1.5 Joules, because whatever is lost by one form of energy is gained by the other.
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RESPONSE --> ok self critique assessment: 2
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