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The velocity of the peg as it moves around the circle is drawn tangentially from the position of the peg. The peg is moving at a constant velocity, v equals omega r equals omega A. Notice that theta-one equals theta-two equals theta-three. Now, sine theta-3 equals minus the velocity along x, v-sub x, divided by the velocity along the wheel, v, and so v sub x equals minus v times sine theta one. By using omega-A equals v, and theta equals omega-t, the equation becomes v sub x equals minus omega-A times sine of omega-t. Try this question: What is the velocity of the spot 1.0 seconds after it has started at its maximum amplitude on the right if the amplitude is 4.8 meters and the wheel rotates at 2.0 radians per second? Pick one of these choices after you have tried to solve the problem. ... Use the velocity formula. ... As the drawing indicates, the position of the spot after 1.0 seconds is at x = -2.0 meters. You may wish to verify this position. Remember that to verify the position, you'll want to use the equation x equals omega-A times cosine omega-t. We have equations for position, x, and velocity, v. The third thing we need to do is to derive an equation for the acceleration, a, of the spot. For this derivation, recall that centripetal acceleration, a-sub-c equals v-squared divided by r, and since v equals omega-r, then v-squared divided by r equals omega-squared times r-squared divided by r, so the centripetal acceleration is omega-squared times r or a-sub-c equals omega-squared times A. As the peg moves at a constant velocity to the top of the circle, the spot on the x axis accelerates. Cosine theta equals negative a-sub-x divided by a-sub-c and a-sub-x equals negative a-sub-c cosine theta or a-sub-x equals negative a-sub-c cosine omega-t. Since the centripetal acceleration, a-sub-c equals omega-squared times A, the acceleration of the spot along the x-axis, a-sub-x equals negative omega-squared times A cosine omega-t. Try this question: What is the acceleration of the spot on the x axis at t = 1.2 seconds? The amplitude is 3.0 meters and the angular velocity, omega, is 0.75 radians/second. Pick one of these choices. ... Notice in the sketch that the spot is 1.9 meters from the equilibrium point. How would you have known that the spot is at that point? The position, 1.9 meters, was obtained by using the equation x equals A cosine-omega-t. After putting in the numbers for the amplitude, the angular velocity and the time, you have x equals 3 times cosine of 0.75 times 1.2 which equals 1.9 meters. Here are the three formulas that describe position, velocity and acceleration of an oscillator. Position, X equals A times cosine omega t. Velocity, v, equals negative omega times A times sine omega t. Acceleration, a, equals negative omega-squared times A times cosine omega t.

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