Suppose that a mass of 100 g stretches a spring 5 cm. If the mass is set in motion from its equilibrium position with a downward velocity of 24.90 cm/sec, and if there is no damping then we can obtain the equation modelling this system as follows: The spring constant is (100g) (980cm/sec²) k 19600. 5cm Hence 100u" +19600u = 0 can be simplified to u" + 196u = 0. In this case the initial values are u(0)=0 and u'(0)=24.90. By using the information given above, determine the value of u(3.2), the position of the mass at 3.2 seconds. (Write the numerical value of the answer into the box) Answer:

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