course phy 201 6 8/2 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.
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Given Solution: `a** 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& 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: a = omega A son (omega * time) centrip acceleration = v^2 /r confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a 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). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& 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: KE = 1/2 k * A^2 - 1/2 k x^2 so 1/2 m v^2 = 1/2 k * A^2 - 1/2 k x^2 v =sqrt (1/2 k/m * A^2 - x^2 confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: we pull back the pendulum and stop it at the equilibrium point, the washer on the pendulum will be projected off and then we can measure the horizontal distance and therefore find the horizontal velocity. confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: 68 kg * 9.8 m/s/s = 670 N 670 N / 0.005 m = 134000 N /m = k velocity = sqrt ( 134000 N/m )/ 1500 kg = 9 rad/sec 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. Your solution: period on mars = sqrt 1/.37 = 1.7 ; 1.7 * .80 sec = 1.3 sec 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 0.37 * 9.8 m/s^2 = 3.6 m/s^2 == new gravity constant period = 2 pi sqrt(L / g) = 2 pi sqrt(.15 m / (3.6 m/s^2)) = 1.3 seconds, confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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). 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. 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: 3 ********************************************* Question: `qQuery gen problem 11.14 80 N to compress popgun spring .2 m with .15 kg ball. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: k = 80 N / .2 = 400 N/m PE = 1/2 * 400 N/m * 0.2^2 KE = 1/2 m * v^2 = 1/2 * 400 N/m * 0.2^2 1/2 * .15 kg * v^2 = 1/2 * 400 * 0.2 ^2 v= 10.3 m/a confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): 3 ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qQuery gen phy problem 11.24 spring 305 N/m amplitude 28 cm suspended mass .260 kg. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: angular frequency = sqrt(305 * .250 / (210 * .260 omega = sqrt ( k/m) = sqrt (210 N/m/ .25kg) = 29 rad/s motion equation = y = 28 cm * sin (29 rad/sec * t) confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: ok