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3 years ago

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A mass is hung from a spring and set in motion so that it oscillates continually up and down. The velocity v of the weight at ti

me t is given by the equation v=−2 cos(3πt) with v measured in feet per second and t measured in seconds. Determine the maximum velocity of the mass and the amount of time it takes for the mass to move from its lowest position to its highest position.

1 answer:

djverab [1.8K]3 years ago

8 0

Answer:

the maximum velocity of the mass v (max) = 2 ft/s

the amount of time it takes for the mass to move from its lowest position to its highest position∆t = 1/3 seconds = 0.33 seconds

Explanation:

Given the velocity equation;

v=−2 cos(3πt)

The maximum velocity would be at cos(3πt) = 1 or cos(3πt) = -1

v (max) = -2 × -1 = 2 ft/s

The time taken for the mass to move from lowest position to highest position

At Lowest position, vertical velocity equals zero.

At highest position, vertical velocity equals zero.

The time taken for the mass to move from one v = 0 to the next v = 0

Cos(π/2) = 0 and

Cos(3π/2) = 0

For the first;

Cos(3πt) = cos(π/2)

3πt1 = π/2

t1 = π/2(3π)

t1 = 1/6 second

For the second;

Cos(3πt) = cos(3π/2)

3πt2 = 3π/2

t2 = 3π/2(3π)

t2 = 1/2 second

∆t = t2 - t1 = 1/2 - 1/6 = 3/6 - 1/6 = 2/6 = 1/3 seconds

∆t = 1/3 seconds

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