Question

A 7 kilogram object suspended from the end of a vertically hanging spring stretches the spring 5 centimeters. At time $t = 0$, the resulting mass-spring system is disturbed from its rest state by the force $F(t) = 133 \cos(14t)$. The force $F(t)$ is expressed in Newtons and is positive in the downward direction, and time is measured in seconds. Determine the spring constant $k$. $k = 1372$ Newtons / meter Hint is to use earth gravity of 9.8 meters per second squared, and note that Newton is kg meter per second squared. Formulate the initial value problem for $x(t)$, where $x(t)$ is the displacement of the object from its equilibrium rest state, measured positive in the downward direction. Give your answer in terms of $x$, $x'$, $x''$, $t$. Differential equation: $x'' + 196x = 19 \cos(14t)$ Initial conditions: $x(0) = 0$ and $x'(0) = 0$ Solve the initial value problem for $x(t)$. $x(t) = $ Plot the solution and determine the maximum displacement from equilibrium made by the object on the time interval $0 \le t < \infty$. If there is no such maximum, enter "NONE". maximum displacement = none meters Book: Section 2.6 of Notes on Diffy Qs

          A 7 kilogram object suspended from the end of a vertically hanging spring stretches the spring 5 centimeters. At time $t = 0$, the resulting mass-spring system is disturbed from its rest state by the force $F(t) = 133 \cos(14t)$. The force $F(t)$ is expressed in Newtons and is positive in the downward direction, and time is measured in seconds.
Determine the spring constant $k$.
$k = 1372$ Newtons / meter
Hint is to use earth gravity of 9.8 meters per second squared, and note that Newton is kg meter per second squared.
Formulate the initial value problem for $x(t)$, where $x(t)$ is the displacement of the object from its equilibrium rest state, measured positive in the downward direction.
Give your answer in terms of $x$, $x'$, $x''$, $t$.
Differential equation: $x'' + 196x = 19 \cos(14t)$
Initial conditions: $x(0) = 0$ and $x'(0) = 0$
Solve the initial value problem for $x(t)$.
$x(t) = $
Plot the solution and determine the maximum displacement from equilibrium made by the object on the time interval $0 \le t < \infty$. If there is no such maximum, enter "NONE".
maximum displacement = none meters
Book: Section 2.6 of Notes on Diffy Qs

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a 7 kilogram object suspended from the end of a vertically hanging spring stretches the spring 5 centimeters at time t 0 the resulting mass spring system is disturbed from its rest state by 92353

Added by Angela L.

Calculus: Early Transcendentals

James Stewart 8th Edition

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Solved by Expert David Mccaslin

07/26/2022

Step 1

The weight of the object is $W = mg = 7(9.8) = 68.6$ N. The spring stretches 5 cm = 0.05 m. Hooke's law states that $F = kx$, where F is the force, k is the spring constant, and x is the displacement. In this case, $F = W = 68.6$ N and $x = 0.05$ m. Therefore, $k Show more…

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A 7 kilogram object suspended from the end of a vertically hanging spring stretches the spring 5 centimeters. At time $t = 0$, the resulting mass-spring system is disturbed from its rest state by the force $F(t) = 133 \cos(14t)$. The force $F(t)$ is expressed in Newtons and is positive in the downward direction, and time is measured in seconds. Determine the spring constant $k$. $k = 1372$ Newtons / meter Hint is to use earth gravity of 9.8 meters per second squared, and note that Newton is kg meter per second squared. Formulate the initial value problem for $x(t)$, where $x(t)$ is the displacement of the object from its equilibrium rest state, measured positive in the downward direction. Give your answer in terms of $x$, $x'$, $x''$, $t$. Differential equation: $x'' + 196x = 19 \cos(14t)$ Initial conditions: $x(0) = 0$ and $x'(0) = 0$ Solve the initial value problem for $x(t)$. $x(t) = $ Plot the solution and determine the maximum displacement from equilibrium made by the object on the time interval $0 \le t < \infty$. If there is no such maximum, enter "NONE". maximum displacement = none meters Book: Section 2.6 of Notes on Diffy Qs

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Transcript

00:01 So in this problem, we have a spring with a 10 kilogram mass on it, and initially it stretches 9 .8 centimeters.

00:06 So we can kind of think about this right.

00:08 Here's a spring that's got a 10 kilogram mass hanging on it here.

00:17 Initially, it stretches when you hang the mass on there.

00:20 Then at time t equals zero, there's some force applied to it, given by f of t is 140 cosine 10t, and forces in newton's, and of course time t is at 7.

00:31 Seconds.

00:33 We're first asked to find the spring constant k.

00:36 Well, initially when you put that weight on there, it stretches and stops, right? so i have k times y, the displacement that it stretches is equal to mass times gravity.

00:50 Well, this displacement right here is 0 .098 meters, like there's 100 centimeters in a meter.

01:00 So this means i have k times 0 .098.

01:04 Is equal to my mass, 10 kilograms, and g, our gravitational pull, is 9 .8 newton's per meter squared, or so per, not meter squared, sorry, nons per kilogram.

01:25 All right, so that means that k is 10 times 9 .8 over kilograms cancel, over 0 .098.

01:36 And so this is 98 over 0 .098, which is 1 ,000 newtons per meter.

01:50 It is my spring gravitational constant.

01:55 My spring constant.

01:57 Okay.

01:59 So, next we're asked to write the initial value problem with conditions.

02:22 Okay.

02:24 Well, if we consider that force is mass times acceleration.

02:31 We have f of t is equal to this 140 cosine 10t, 140 cosine of 10t, minus ky, right, my spring constant times the displacement on it.

02:58 Okay, so that's the sum of the forces.

03:00 And that's equal to mass times acceleration.

03:02 Well, acceleration is the second derivative of displacement with respect to time, which i could write this as m y double prime.

03:12 And we know k is a thousand.

03:14 So that means i have 140 cosine of 10 t is equal to my mass is 10, 10 y double prime, plus a thousand y.

03:39 I can divide everything here by 10...

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