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An object of mass m = 40 grams attached to a coiled spring with damping factor b = 0.75 gram/second is pulled down a distance a = 15 centimeters from its rest position and then released. Assume that the positive direction of the motion is up and the period is T = 5 seconds under simple harmonic motion.

(a) Develop a model that relates the distance d of the object from its rest position after t seconds.

*(b) Graph the equation found in part (a) for 5 oscillations.

Short Answer

Expert verified

The equation isd(t)=15e-0.75t/80cos4Ï€252-0.7521602t

The graph of the motion is

Step by step solution

01

Part (a) Step 1. Given Information

  • The mass of the object is 40 grams.
  • Distance from rest position is 15 centimeters.
  • Damping Factor is 0.75gram/sec.
  • The time period is 5 seconds.
02

Part (a) Step 2. Develop the model

  • It is given that m=40,a=15,b=0.75,t=5.
  • As the period is 5 seconds, so, 2Ï€Ó¬=5Ó¬=2Ï€5
  • Substitute the values into the equation dt=ae-bt/2mcosÓ¬2-b24m2tand simplify.

dt=15e-0.75t/80cos2Ï€52-0.7521602t=15e-0.75t/80cos4Ï€252-0.7521602t

03

Part (b) Step 1. Given Information

  • The mass of the object is 40 grams.
  • Distance from rest position is 15 centimeters.
  • Damping Factor is 0.75gram/sec.
  • The time period is 5 seconds.
04

Part (b) Step 2. Plot the Graph

Use the obtained equation to plot the graph of the function.

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