Example13.3 - Example 13.3 PHY231, Spring 2008 A 36 kg...

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Example 13.3 PHY231, Spring 2008 A 36 kg block is attached to a spring of constant k = 600 N/m. The block is pulled 3.5 cm away from its equilibrium position and is pushed so that it has an initial velocity of 5.0 cm/s at t = 0. What is the position of the block at t = 0 . 75 seconds? Answer: The formula for position as a function of time is: x ( t ) = A cos( ωt - φ ) . We will also need velocity as a function of time: v ( t ) = - ωA sin( ωt - φ ). We are given: m = 36 kg, k = 600 N/m, x (0) = 3 . 5 cm, v (0) = 5 . 0 cm/s. Note that the amplitude is not 3.5 cm, because it does not start from rest. The initial push will make the amplitude larger than 3.5 cm. First, we can ±nd the angular frequency from: ω = r k/m = 4 . 0825 rad / s . Next, we must ±nd the amplitude and the phase. In general, A and φ are always determined by the initial conditions, x (0) and v (0). We plug t = 0 into the equations for position and velocity and set them equal to the initial conditions: x (0) =
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This note was uploaded on 07/25/2008 for the course PHY 231 taught by Professor Smith during the Spring '08 term at Michigan State University.

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Example13.3 - Example 13.3 PHY231, Spring 2008 A 36 kg...

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