We found that the time after the mass passes through the equilibrium point at which the mass attains its extreme displacement from the equilibrium position is 3 4 seconds. Lastly, we must find the position of the mass at the instant the mass attains its extreme displacement from the equilibrium, which is x 3 4 . We note that the result is measured in feet. x(t) = −e−4t + 2te−4t

A) X(0.5) = 0 ft

B) X(0.75) = 0.023 ft.

Explanation: Given that the result is measured in feet.

x(t) = −e−4t + 2te−4t

Factorizing the above equation lead to

x(t) = e^-4t( -1 + 2t )

The mass passes through the equilibrium position when x(t) = 0

0 = e^-4t( -1 + 2t )

-1 + 2t = 0

2t = 1

t = 0.5s

x(t) = e^-4t( -1 + 2t )

Substitute t = 0.5

x(0.5) = e^-4(0.5)(-1 + 2(0.5))

X(0.5) = 0

Also given that mass attains its extreme displacement from the equilibrium position is t = 3/ 4 

= 0.75 seconds

x(t) = e^-4t( -1 + 2t )

x(t) = e^-4(0.75)( -1 + 2(0.75) )

X(0.75) = (e^-3)(0.5)

X(0.75) = 0.023 ft.

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