Question

The displacement from equilibrium of an oscillating weight suspended by a spring and subject to the damping effect of friction is given by y(t) = 2e−t cos 4t, where y is the displacement (in centimeters) and t is the time (in seconds). Find the displacement when t = 0, t = 1 4 , and t = 1 2 . (Round your answers to two decimal places.)

Answer 1

The displacement from equilibrium is given by y(t) = 2e-t cos 4t. When t = 0, y = 2 cm. When t = 1 4 , y = 0.315 cm. When t = 1 2 , y = -0.493 cm.

Step-by-step explanation:

The displacement from equilibrium of an oscillating weight suspended by a spring and subject to the damping effect of friction is given by the equation y(t) = 2e-tcos(4t), where y is the displacement (in centimeters) and t is the time (in seconds).

To find the displacement when t = 0, t = 1/4, and t = 1/2, we substitute the values of t into the equation.

When t = 0, y(0) = 2e0cos(0) = 2(1)(1) = 2 centimeters.

When t = 1/4, y(1/4) = 2e-1/4cos(1) = 2(0.7788)(0.2027) = 0.315 centimeters.

When t = 1/2, y(1/2) = 2e-1/2cos(2) = 2(0.6065)(-0.4161) = -0.493 centimeters.

Answer 2

Note that the correct times are t = 0, t = 1/4, t = 1/2. You can tell from the spaces between the two digits. i.e 1&4 and 1&2

Answer:

y(0) = 2.00 cm

y(1/4) = 1.56 cm

y(1/2) = 1.21 cm

Step-by-step explanation:

This is a very simple exercise, the displacement of the oscillating weight from equilibrium has already been modeled by the equation:

`y(t) = 2e^(-t) cos 4t`

Where y = displacement ( in cm)

and t = time (in seconds)

The task is to find the displacement when t = 0, 1/4 and 1/2

When t = 0 s

`y(0) = 2e^(0) cos 4(0)\\y(0) = 2* 1*1\\y(0) = 2.00 cm`

When t = 1/4 s

`y(1/4) = 2e^(-1/4) cos 4(1/4)\\y(14) = 2e^(-1/4) cos (1)\\y(1/4) = 1.56 cm`

When t = 1/2

`y(1/2) = 2e^(-1/2) cos 4(1/2)\\y(14) = 2e^(-1/2) cos (2)\\y(1/2) = 1.21 cm`