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Question 1 (35 pts.): Consider the spring-mass-damper system as shown below. The initial displacement is \(-0.1\) m and the initial velocity is zero. No external force is applied. At time $t = 3$ seconds, the spring is disconnected. Find the position and velocity of the mass at $t = 5$ seconds. $M = 1$ kg, $f_v = 0.5$ N/m/s, $K = 0.5$ N/m, $f(t) = 0$ N, $x(0) = -0.1$ m

          Question 1 (35 pts.): Consider the spring-mass-damper system as shown below. The initial displacement is \(-0.1\) m and the initial velocity is zero. No external force is applied. At time $t = 3$ seconds, the spring is disconnected. Find the position and velocity of the mass at $t = 5$ seconds.
$M = 1$ kg, $f_v = 0.5$ N/m/s, $K = 0.5$ N/m, $f(t) = 0$ N, $x(0) = -0.1$ m
        
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Question 1 (35 pts.): Consider the spring-mass-damper system as shown below. The initial displacement is -0.1 m and the initial velocity is zero. No external force is applied. At time t = 3 seconds, the spring is disconnected. Find the position and velocity of the mass at t = 5 seconds.
M = 1 kg, fv = 0.5 N/m/s, K = 0.5 N/m, f(t) = 0 N, x(0) = -0.1 m

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University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
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Consider the spring-mass-damper system as shown below. The initial displacement is -0.1 m and the initial velocity is zero. No external force is applied. At time t = 3 seconds, the spring is disconnected. Find the position and velocity of the mass at t = 5 seconds. M = 1 kg, fv = 0.5 N/m/s, K = 0.5 N/m, ft = 0 N, x0 = -0.1 m -.x(1) K 0000 M f(1 fv
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Transcript

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0:00 Question.
00:01 You're giving a damper.
00:03 Basically is a damped harmonic oscillator with mass m.
00:10 And obviously the equation motion for this damp time oscillator is m times acceleration equals the force.
00:20 And you have three forces.
00:22 One is they recall your first, which is simply minus k times x.
00:26 K is brain constant.
00:28 And you have the first, uh, the damping first, which is minus b, b is the dammit coefficient times the velocity, x.
00:37 Dot.
00:37 And of course, you also have the first external force, which is i'll write as f, um, n, sine, omega, t.
00:53 So, um, all you can rewrite this equation as, you can realize this equation as, um, x dot divided by m on both sides, plus k over m x plus b over m x.
01:15 Dot equals f0 sine omega t.
01:24 Now the solution to this equation has two parts.
01:35 The first part, i'll write the solution as xt.
01:39 The first part is the part that solves the homogeneous part of this equation.
01:45 And that part, you can always write that homogeneous part, you can always write as a to the power of b over 2m times t...
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