Amass weighing 1 slug is attached to the end of a spring with spring constant 26 lb/ft. damping force is numerically equal to 2 times the velocity. the mass is pulled down 6 ft and then released from rest. write down and solve the initial value problem for the position x(t)

x = e^(-t) [1.2 sin(5t) + 6 cos(5t)]

Step-by-step explanation:

A damped oscillator has the equation of motion:

m d²x/dt² + β dx/dt + k x = 0

This is an example of a second order linear ordinary differential equation with constant coefficients.

d²x/dt² + b dx/dt + c x = 0

Notice the leading coefficient is 1.

x = C₁ e^(-½ t (b + √(b²−4c) )) + C₂ e^(-½ t (b − √(b²−4c) ))

x = (C₁ t + C₂) e^(-bt/2)

x = e^(-bt/2) [C₁ sin(½ t √(4c−b²)) + C₂ cos(½ t √(4c−b²))]

Given that m = 1, β = 2, and k = 26:

d²x/dt² + 2 dx/dt + 26 x = 0

Here, b = 2 and c = 26, so:

b²−4c = (2)²−4(26) = -100 < 0

The general solution is:

x = e^(-2t/2) [C₁ sin(½ t √100) + C₂ cos(½ t √100)]

x = e^(-t) [C₁ sin(5t) + C₂ cos(5t)]

To find the values of C₁ and C₂, first find dx/dt, then plug in the initial conditions.

dx/dt = e^(-t) [5C₁ cos(5t) − 5C₂ sin(5t)] − e^(-t) [C₁ sin(5t) + C₂ cos(5t)]

dx/dt = e^(-t) [5C₁ cos(5t) − 5C₂ sin(5t) − C₁ sin(5t) − C₂ cos(5t)]

dx/dt = e^(-t) [(5C₁ − C₂) cos(5t) − (5C₂ + C₁) sin(5t)]

Given x(0) = 6:

6 = e^(0) [C₁ sin(0) + C₂ cos(0)]

6 = C₂

Given x'(0) = 0:

0 = e^(0) [(5C₁ − C₂) cos(0) − (5C₂ + C₁) sin(0)]

0 = 5C₁ − C₂

0 = 5C₁ − 6

C₁ = 1.2

So the solution is:

x = e^(-t) [1.2 sin(5t) + 6 cos(5t)]

Here's the graph:

desmos.com/calculator/bavfsoju5c

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step-by-step explanation:

on a coordinate grid, point a is at (−4.0, −2.4) and point b is at (4.0, −2.4). point b is a reflection of point a across the –axis. input either a lowercase x or i think is the right awnser

step-by-step explanation: