#$&* course Phy 231 8/4 1 036. `query 36*********************************************
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Given Solution: `a** Position at clock time is x = Acos(`omega* t) Velocity = -`omega *A*sin(`omega* t) Accel = -`omega^2 * A * cos(`omega* t) University Physics students should note that velocity and acceleration are the first and second derivatives of the position function. ** The acceleration should be -omega^2 A cos(omega * t). The velocity function is the derivative of the position function with respect to clock time. That function is a composite of the function A cos(z) with z = omega * t. The derivative of A cos(z) is - A sin(z) and the derivative of omega * t is just omega, so the derivative of the composite is v(t) = x ' ( t) = (omega * t) ' * (-A sin(omega * t) ) = - omega A sin(omega t).. The acceleration function is the derivative of the velocity function so again using the chain rule we obtain a(t) = - omega^2 A cos(omega t). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&* OK ********************************************* Question: `qHow is the acceleration of the pendulum related to the centripetal acceleration of the point on the reference circle? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: When the pendulum is at the equilibrium point the centripetal acceleration of the point and the acceleration of the pendulum will be equal. This is when the accelerations are going in the same direction. Cent accel = v^2 / r Acceleration = - omega^2 A cos (omega t) confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: a = -`omega A sin(`omega *t) and aCent = v^2/r for the circle modeling SHM ** The centripetal acceleration of the point on the reference circle, which acts toward the center of the circle, has two components, one in the x direction and one in the y direction. The component of the centripetal acceleration in the direction of the motion of the oscillator is equal to the acceleration of the oscillator. If the oscillator is at position theta then the centripetal acceleration has direction -theta (back toward the center of the circle, opposite to the position vector). The centripetal acceleration is aCent = v^2 / r; so the x and y components are respectively ax = aCent * cos(-theta) = v^2 / r * cos(theta) and ay = aCent * sin(-theta) = -v^2 / r * sin(theta). ** The reference circle picture indicates a radial position vector following the reference point around the circle of radius A at angular velocity omega. The reference point has a velocity vector tangent to the circle with magnitude omega * A, and a centripetal acceleration vector directed toward the center of the circle with magnitude v^2 / r = (omega * A)^2 / A = omega^2 * A. In this example we are modeling the motion of an oscillator moving along the x axis. The x components of the position, velocity and acceleration vectors are A cos(omega t), -omega A sin(omega t) and -omega^2 cos(omega t). Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&* OK 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: PE = .5 k A^2 Total energy = .5 k x^2 PE + KE = total energy .5 k A^2 + KE = .5 k x^2 .5 m v^2 = .5 k x^2 - .5 k A^2 V = (.5 k / m (x^2 – A^2) confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** The PE of the pendulum at displacement x is .5 k x^2. By conservation of energy, if nonconservative forces are negligible, we find that the KE of the pendulum at position x is.5 k A^2 - .5 k x^2. This result is obtained from the fact that at max displacement A the KE is zero, and the KE change from displacement A to displacement x is the negative of the PE change between these points. Thus .5 m v^2 = .5 k A^2 - .5 k x^2. Solving for v we have v = +- sqrt( .5 k / m * (A^2 - x^2) ) . ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&* OK ********************************************* Question: `qHow can we determine the maximum velocity of a pendulum using a washer and a rigid barrier? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The max velocity happens at the equilibrium position. You pull back the pendulum to a small distance and let it go. Put the barrier at the equilibrium point and make it so that the washer will fall off the pendulum at that point. It will cover a horizontal distance like a projectile, so you can use the vertical distance and the time to determine the horizontal velocity of the washer, which will be the maximum velocity of the pendulum. confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aGOOD STUDENT ANSWER: If we pullback a pendulum of length L a distance x (much smaller than L), and stop the motion at the equilibrium point (vertical limit of motion) a washer on the pendulum will become a projectile and project off the pendulum, to land at a distance from which we can determine the horizontal velocity of the washer. That velocity is the same as the max velocity of the pendulum, since the max velocity is that which is at the lowest point in its path. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&* OK ********************************************* 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: Weight of the person = 666 N Force constant = 666N / .005 m = 133 200 N/m Angular frequency = sqrt (133200 N/m / 1568kg) = 9.22 rad/s 9.22 rad/s / 2pi = 1.47 cycles / second confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aFrom the weight of the driver and the compression of the spring, we determine the spring constant (the 'stiffness' of the spring in N / m): driver weight of 68 kg * 9.8 m/s^2 = 670 N compresses the spring .05 meters, so since | F | = k | x | we have k = | F | / | x | = 670 N / (.005 m) = 134 000 N / m. Now from the force constant and the mass of the system we have omega = sqrt(k / m) = sqrt( (134 000 N/m) / (1570 kg) ) = 9 sqrt( (N/m) / kg) ) = 9 sqrt( (kg / s^2) / kg) = 9 rad / s, approximately, or about 1.5 cycles / second. Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&* OK ********************************************* 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: Omega = sqrt (g / L) – this means that the period is proportional to the square root of the gravitational force Sqrt(1 / .37) = 1.64 which is the proportionality of the accelerations due to gravity Period on mars = 1.64 * .8 s = 1.31 seconds confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe period of a angular frequency harmonic oscillator is sqrt(k / m), and the time required for a cycle, i.e., the period of the cycle, is the time required to complete a cycle of 2 pi radians. For a pendulum we have k = sqrt( m g / L ), where g is the acceleration of gravity. Thus for a pendulum omega = sqrt(k / m) = sqrt( (m g / L) / m) = sqrt( g / L). From this we see that for a given length, the frequency of the pendulum is proportional to sqrt(g). The period is inversely proportional to the frequency, so the period is inversely proportional to sqrt(g). Thus we have period on Mars / period on Earth = sqrt( gravitational acceleration on Earth / gravitational acceleration on Mars) = sqrt( 1 / .37) = 1.7, approximately. So the period on Mars would be about 1.7 * .80 sec = 1.3 sec, approx. As an alternative to the reasoning or proportionality, we can actually determine the length of the pendulum, and use this length with the actual acceleration of gravity on Mars. We have period = 2 pi rad / angular frequency = 2 pi rad / (sqrt( g / L) ) = 2 pi rad * sqrt(L / g). We know the period and acceleration of gravity on Earth, so we can solve for the length: Starting with period = 2 pi sqrt(L / g)) we square both sides to get period^2 = 4 pi^2 L / g. Multiplying both sides by g / (4 pi^2) we get L = g * period^2 / (4 pi^2) = 9.8 m/s^2 * (0.80 sec)^2 / (4 pi^2) = .15 meters. The pendulum is .15 meters, or 15 cm, long. On Mars the acceleration of gravity is about 0.37 * 9.8 m/s^2 = 3.6 m/s^2, approx.. The period of a pendulum on Mars would therefore be period = 2 pi sqrt(L / g) = 2 pi sqrt(.15 m / (3.6 m/s^2)) = 1.3 seconds, approx. This agrees with the 1.3 second result from the proportionality argument. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&* OK ********************************************* Question: `qQuery gen problem 11.14 80 N to compress popgun spring .2 m with .15 kg ball. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: K = 80N / .2m = 400 N/m PE = (1/2)(400N/m)(.2m)^2 = 8 J 8J = (1/2)(.15kg)(v^2) V = 10.3 m/s confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The PE of the system will be .5 k A^2, where A = .2 m and k = F / x = 80 N / (.2 m) = 400 N / m. The KE of the released ball will in the ideal case, which is assumed here, be .5 m v^2 = .5 k A^2. Solving for v we obtain v = +- sqrt( k A^2 / m ) = +- sqrt( 400 N/m * (.2 m)^2 / (.15 kg) ) = +- sqrt( 106 m^2 / s^2) = +-10.3 m/s, approx. The speed of the ball is the magnitude 10.3 m/s of the velocity. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&* OK ********************************************* Question: `qQuery gen phy problem 11.24 spring 305 N/m amplitude 28 cm suspended mass .260 kg. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Omega = sqrt(305N/m / .26 kg) = 34.3 rad/s Equation of motion: y = .28m sin (34.3 rad/s * t + theta0) confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The angular frequency of the oscillation (the angular velocity of the point on the reference circle) is omega = sqrt(k / m), with k = 210 N/m and m = .250 kg. The equation of motion could be y = A sin(omega * t). We obtain omega = sqrt( 210 N/m / (.250 kg) ) = sqrt( 840 s^-2) = 29 rad/s, approx.. A is the amplitude 28 cm of motion. So the equation could be y = 28 cm sin(29 rad/s * t). The motion could also be modeled by the function 28 sin (29 rad/s * t + theta0) for any theta0. The same expression with cosine instead of sine would be equally valid, though for any given situation theta0 will be different for the cosine model than for the sine model. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&* OK ********************************************* 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: K = 40N / .25m = 160 N/m Omega = 1s * 2pi = 2pi rad/s 2pi = sqrt (160N/m / M) M = 4.05 kg A = .05m Position = .05 sin (2pi rad/s* .35s) = .05 sin(2.199rad) = -.0405 meters from equilibrium (it was going down) Velocity = -2pi * .05 cos (2pi rad/s * .35s) = -.185 meters/s confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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 ** 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.. ** 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). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&* OK ********************************************* Question: `qQuery Add comments on any surprises or insights you experienced as a result of this assignment. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): No big surprises, I don’t think… ------------------------------------------------ Self-critique rating #$&* OK "