A spring with a 8-kg mass and a damping constant 8 can be held stretched 2 meter

Question

A spring with a 8-kg mass and a damping constant 8 can be held stretched 2 meters beyond its natural length by a force of 8 newtons. Suppose the spring is stretched 4 meters beyond its natural length and then released with zero velocity, In the notation of the text, what is the value c2?4mk?______________ Find the position of the mass after t seconds. Your answer should be a function of the variable t with the general form c1e^(?t)cos(?t)+c2e^(?t)sin(?t) ?=________________ ?=______________ ?= ______________ ?= _____________ c1=____________ c2= ______________

Explanation / Answer

"I can take derivatives and calculate constants just fine, but the biggest issue is setting up the equation." Of course! This is the hard part. "The textbook's examples are not even close to this one," Someday your textbooks will not have any "examples" at all, and you must develop sufficient competency in order to use first principles for specific problems. Just a heads up. I just always hear students say how they can do blind math, but all they need is to set up the problem. This is expected, the hard part *is* setting up the problem, the math is never the hard part. Anyway, m = 9 kg, k = 11 N/m, The first step in any physics problem is to choose a coordinate system, I will choose on that is measured at the equilibrium position (when the mass is just hanging there). So that y measures vertical motion, y = 0 is the equilibrium position, downward is *positive*, upward is negative by choice. An equation is generated from Newton's second law: Sum of forces in the y direction = mg - F_s = ma_y .............Eqn. (1) subscript y means acceleration in the y direction, F_s = spring force, mg = force due to gravity. Now, a_y = d^2 y / dt^2 since y is measured from the equilibrium position, F_s = F_s0 + ky F_s0 = equilibrium spring force, analyze the equilibrium position. Equilibrium happens if the sum of forces = 0, Equilibrium: mg - F_s0 = 0 (recall downward is positive) This gives F_s0 = mg, thus F_s = mg + ky, and ky comes from the spring equation (Hooke's law), it is the additional restoring force that is incurred when you displace the mass away from its equilibrium position. Note that y = 0 is the equilibrium position, and downward is positive by our choice. If you stretch the spring further down, y is positive (downward), the spring tries to restore itself to by pulling upwards (negative), so this actually makes F_s positive. This is exactly why in Eqn. (1) it is written as mg - F_s = ma_y there is a minus sign in front of F_s so as to give it the proper sign. You can convince yourself that if you push the mass up ( y is negative), you get the right sign of the force in that equation (the net effect is for the spring to pull it back down, i.e. -F_s is positive). Let us put all we know into Eqn. (1): mg - F_s = ma_y mg - (mg + ky) = m d^2 y /dt^2 -ky = m d^2 y/dt^2 d^2 y/dt^2 + (k/m)y = 0 This is our governing equation, and notice it is the same equation you get if the spring is being stretched horizontally on say, the surface of a table, where gravity is not even acting in the same direction as the spring motion. Can you explain the meaning of c^2 - 4mk? Symbols are not universal, and ordinarily people use symbols like m = mass, k = spring constant, but c? What is that? I am not in your class, so I do not know what particular notation your teacher/professor/textbook uses as it is not standard. To do the second part, I am going to ignore what your form is, and just find the solution and show it matches. Solve d^2 y/dt^2 + (k/m)y = 0 let w^2 = k/m (note, I am squaring it because we will have to take a square root later, this is just for convenient bookkeeping, if you like you could make w = k/m, there is no difference). d^2 y/dt^2 + w^2 y = 0 This is a second order ordinary linear differential equation with constant coefficients, this diagnosis informs that the solutions may be seeked in the form y ~ e^{rt}, for values of r to be determined by the differential equation. Model y as e^{rt}, put it in the equation, and see what values of r work: y ~ e^{rt} d^2y/dt^2 ~ r^2 e^{rt} then the differential equation gives r^2 e^{rt} + w^2 e^{rt} = 0 e^{rt} { r^2 + w^2} = 0 e^{rt} cannot be zero, this means that for the equation to work: r^2 + w^2 = 0 r^2 = -w^2 r = +/- i w so two values of r work, r = iw, and r = -iw, thus the general solution (we said the solution y looked like e^{rt}) is y(t) = Ae^{i w t} + B e^{-i w t}, where A, B = constants to be determined Recall e^{i w t} = cos (wt) + i sin(wt), then y(t) = A{ cos (wt) + i sin(wt) } + B{ cos(wt) - i sin(wt)} = (A + B) cos (wt) + i(A - B) sin (wt) y(t) = D cos (wt) + E sin (wt), where D = A + B = constant, E = i(A - B) = constant,the labels have just been changed for convenience. Find the constants, let t = 0 be the moment of release when it is at y = 2 m, and released with velocity v = 0. Assemble two equations: y(0) = 2 m = D Get the second from the velocity = dy/dt dy/dt = w{ E cos (wt) - D sin (wt) } dy/dt|_{t = 0} = 0 = wE --> E = 0 Thus, y(t) = (2 m) cos (w t) Calculate w = sqrt{k/m} = sqrt{11/9} [1/sec] so y(t) = (2 m) cos{ sqrt{11/9} t }................[Ans.] This matches the general form given for values:---------------------- C1 = 2 m alpha = 0 Beta = w = sqrt{11/9} C2 = 0 lambda = 0 omega = 0

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