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Unformatted text preview: Version 093 – TEST04 – TSOI – (58160) 1 This printout should have 15 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points The figure below shows a complex wave pat tern on a string moving towards a rigid hook at the wall on the right. After some time, the wave is reflected from the wall. v Select the wave pattern for the reflected wave. 1. v 2. v correct 3. 4. v 5. v Explanation: Consider the wave pattern image reflected about the rigid hook on the wall. v v After the time it takes for the wave to be reflected from the wall, this image is the wave pattern traveling to the left along the string. Note: Reflection about a point (hook) is the same as reflection about the yaxis (wall) fol lowed by reflection about the xaxis (string). The leading part of the wave must remain in front and the wave is flipped over. 002 10.0 points Hint: Consider a uniform rod with a mass m and length L pivoted on a frictionless horizontal bearing at a point O parenleftbigg 2 3 L from the lower end parenrightbigg , as shown in the figure. 2 3 L L O θ What is the period of this pendulum? Use the small angle approximation. The moment of inertia of a uniform rod about its center of mass is 1 12 M L 2 . 1. T = 2 π radicalBigg 14 15 L g 2. T = 2 π radicalBigg 14 9 L g 3. T = 2 π radicalBigg 7 12 L g 4. T = 2 π radicalBigg 2 3 L g correct Version 093 – TEST04 – TSOI – (58160) 2 5. T = 2 π radicalBigg 62 105 L g 6. T = 2 π radicalBigg 38 63 L g 7. T = 2 π radicalBigg 19 24 L g 8. T = 2 π radicalBigg 43 72 L g 9. T = 2 π radicalBigg 26 21 L g 10. T = 2 π radicalBigg 26 45 L g Explanation: Using the parallel axis theorem, the mo mentum of inertia is I ≡ M d 2 . In this case there are two masses with I O = I cm + M D 2 , where D = parenleftbigg 2 3 − 1 2 parenrightbigg L = 1 6 L , so (1) I O = M bracketleftBigg 1 12 + parenleftbigg 1 6 parenrightbigg 2 bracketrightBigg L 2 = 1 9 M L 2 . (2) When the rod is at an angle θ with respect to the vertical direction, the lever arm due to the weight about point O is D sin θ ≈ D θ using the small angle approximation. In turn the torque due to the weight about point O is given by the equation of motion τ = Iα, so I O d 2 θ dt 2 = − mg D θ . D is given in Eq. 1 and I O is given in Eq. 2, and for simpleharmonicmotion, we have d 2 θ dt 2 = − ω 2 θ = − M g D I O θ = − 1 6 M g L 1 9 M L 2 θ = − 3 2 g L θ . (3) The expression for simple harmonic motion and Eqs. 1 and 2, yields ω 2 = 2 M g D I O , so T ≡ 2 π ω = 2 π radicalBigg I O 2 M g D = 2 π radicaltp radicalvertex radicalvertex radicalvertex radicalvertex radicalvertex radicalbt parenleftbigg 1 9 parenrightbigg M L 2 parenleftbigg 1 6 parenrightbigg M g L = 2 π radicalBigg 2 3 L g . (4) keywords: 003 10.0 points A liquid of density 1123 kg / m 3 flows with speed 1 . 24 m / s into a pipe of diameter 0 . 18 m ....
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This note was uploaded on 11/21/2010 for the course PHY 58160 taught by Professor Tsio during the Fall '10 term at University of Texas.
 Fall '10
 TSIO

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