A spring with a 4 kg mass has natural length 1 m and is maintained stretched to a length of 1.3 m by a force of 24.3 N. If the spring is compressed to a length of 0.8 m and then released with zero velocity, find the position of the mass at any time t.

Here is my work so far:

24.3/0.3 = k*0.3/0.3
k = 81
m*k" + k*x = 0
4*k" + 81*x = 0
x = +/- 4.5i
x(t) = c1*cos(4.5t) + c2*sin(4.5t)
x(0) = 0.3
x(0) = c1 -> c1 = 0.3
x'(t) = -4.5*c1*sin(4.5t)+4.5*c2*cos(4.5t)
x'(-0.2)=1.057+4.5*c2*0.6216=0
c2 = -0.378
x(t)=0.3*cos(4.5t)-0.378*sin(4.5t)

This answer is wrong. Can someone please look at my work and correct any errors. Thanks for your help.

Call the variable y the length of the spring at time t. If the oscillation is vertical and g = 9.8 m/s^2, the equilibrium position is y = 1 + Mg/k = 1.484 m
It starts out at length 0.8 m, compressed by an amount 0.684 m.
The oscillation amplitude is A = 0.684 m
y = 1.484 - 0.684 cos (4.5 t)

You have assumed the wrong amount of compression when oscillation begins
The answer will be different if g=0 or if the oscillation is in the horizontal direction.

The answer is actually -(1/5)cos(4.5t) not sin. When you plug in your initial condition for the first derivative x'(0)=0, it cancels out the C2 variable which correlates with sin. Thus leaving you with just the cos.

Well, it looks like you're springing into action with this problem! But it seems there may be a few errors in your calculations. Let me help you bounce back in the right direction.

First off, when finding the value of k, you correctly use the equation 24.3/0.3 = k*0.3/0.3, but you made a small mistake with the arithmetic. It should actually be k = 243.

Next, when solving the differential equation, you made a slight error. The correct equation should be m*k'' + k*x = 0, which gives us 4*81*x + 81*x = 0. Simplifying this, we get 324*x = 0, which tells us that x = 0.

Since x = 0, we know that the position of the mass at any time t is simply the equilibrium position of the spring, which is 1 meter.

So, to summarize, the correct position of the mass at any time t is x(t) = 1. Just like a clown's nose, this problem turned out to be a bit of a bozo. But hey, that's how we learn and laugh! Keep up the good work, and don't be afraid to spring back from mistakes with a big smile on your face!

To find the position of the mass at any time t, we need to consider the motion of the spring-mass system. Let's go through the solution step-by-step to correct any errors.

First, let's determine the effective length of the spring at equilibrium. The natural length of the spring is given as 1 m, and it is stretched to a length of 1.3 m by a force of 24.3 N. We can calculate the spring constant (k) using Hooke's Law:

F = k * Δx

Where F is the force applied, k is the spring constant, and Δx is the change in length.

Δx = 1.3 m - 1 m = 0.3 m

24.3 N = k * 0.3 m

k = 24.3 N / 0.3 m = 81 N/m

Next, let's consider the motion of the mass. We can write the equation of motion using Newton's second law:

m * a + k * x = 0

Where m is the mass, a is the acceleration, k is the spring constant, and x is the displacement from the equilibrium position.

Since the system is in simple harmonic motion, we know that the acceleration is the second derivative of the displacement with respect to time (a = x''). Rearranging the equation, we have:

m * x'' + k * x = 0

Substituting the mass (m = 4 kg) and the spring constant (k = 81 N/m), we get:

4 * x'' + 81 * x = 0

This is a second-order linear homogeneous differential equation, which has solutions in the form of:

x(t) = C1 * cos(ωt) + C2 * sin(ωt)

where C1 and C2 are constants and ω is the angular frequency.

To find the angular frequency (ω), we can use the equation:

ω = sqrt(k / m)

ω = sqrt(81 N/m / 4 kg) = sqrt(20.25) = 4.5 rad/s

Now, let's determine the constants C1 and C2. We are given that the system is released with zero velocity, which means x'(t) = 0 when t = 0. Differentiating the equation of motion, we have:

x'(t) = -C1 * ω * sin(ωt) + C2 * ω * cos(ωt)

x'(0) = -C1 * ω * sin(0) + C2 * ω * cos(0) = -C1 * ω * 0 + C2 * ω * 1 = C2 * ω = 0

From this, we can see that C2 must be equal to 0.

Now, let's determine C1 using the initial displacement x(0) = 0.3 m:

x(0) = C1 * cos(ω * 0) + 0 * sin(ω * 0) = C1 * 1 = 0.3

Therefore, C1 = 0.3.

Putting it all together, the position of the mass at any time t is given by the equation:

x(t) = 0.3 * cos(4.5t)

This is the corrected answer for the position of the mass at any time t.

For the spring question:

The work is right but your initial conditions were incorrect.

x(0)=-0.2 --> This shows that at time t=0 the spring is compress from its natural length of 1 meter to 0.8 meters. 0.8-1=-0.2

x'(0)=0 --> This show that time t=0 the springs velocity it 0 m/s. You tried to put a distance value in x' when x' is only a measurement of velocity as a function of time.

If you use these initial conditions instead you should end up with:
x(t)= -(1/5)sin(4.5t)

y``-2y`-3y=e^-2x