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Example13.3

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

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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 find the angular frequency from: ω = radicalBig k/m = 4 . 0825 rad / s . Next, we must find 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) = A cos( - φ ) = A cos( φ ) = 3 . 5 cm . v (0) = - ωA sin( - φ ) = ωA sin( φ ) = 5 . 0 cm / s . Note that I used cos( - φ ) = cos( φ ) and sin( - φ ) = - sin( φ ). We now have 2 equations

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