Question

The cylinder is displaced 0.17 m downward from its equilibrium position and is released at time t = 0. Determine the displacement y and the velocity v when t = 3.1 s. The displacement and velocity are positive if downward, negative if upward. What is the magnitude of the maximum acceleration?

Answer 1

Complete Question

The image of this question is shown on the first uploaded image

Answer:

a

`d =0.161 \ m`

b

`v = - 0.054 \ m/s`

c

`a = 6.12 \ m/s^2`

Step-by-step explanation:

From the question we are told that

The maximum displacement is A = 0.17 m

The time considered is
`t = 3.1 \ s`

The spring constant is
`k = 137 \ N \cdot m`

The mass is
`m = 3.8 \ kg`

Generally given that the motion which the cylinder is undergoing is a simple harmonic motion , then the displacement is mathematically represented as

`d = A cos (w t )`

Where
`w` is the angular frequency which is mathematically evaluated as

`w = \sqrt{(k)/(m) }`

substituting values

`w = \sqrt{(137)/( 3.8) }`

`w =6`

So the displacement is at t

`d = 0.17 cos (6 * 3.1 )`

`d =0.161 \ m`

Generally the velocity of a SHM(simple harmonic motion) is mathematically represented as

`v = - Asin (wt)`

substituting values

`v = - 0.17 sin ( 6 * 3.1 )`

`v = - 0.054 \ m/s`

Generally the maximum acceleration is mathematically represented as

`a = w^2 * A`

substituting values

`a_(max) = 6^2 * (0.17)`

substituting values

`a = 6^2 * (0.17)`

`a = 6.12 \ m/s^2`