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3. Defining the Total Potential Energy. We know that when a force is conservative, we can calculate the work done by that force W = and define the potential energy according to the equations: 7. Measuring What Really Happens. Understanding the Mass on a Spring System. Figure 1: Mass on a Spring, At Rest W = S?F(x) dx (6) APE = -W (7) When an actual experiment was set up to measure the behavior of a mass on a spring (as An ideal massless and frictionless spring APE = - SX? F(x) dx 8) commonly done in PHYS-120) the following data was found: is assumed have no mass and no friction, and the force obeys the following equation: Measured x1 = -0.177 m Measured F1 = 4.9 N APE = PE = Measured X2 = -0.549 m Measured F2 = 14.7 N M = 1 kg F = -kx (1) 4 x y = 0 here Ms = 169.15 g Vmax = 0.308 m/ max = 0.061 n 4. Including the Mass of the Spring. amax = 1.549 m/$2 T = 1.246 s Where k is the spring constant measured in N/m, and y is the extension of the spring from +y dir Stretched by yo and In practice, all springs have mass. We will use a model for a spring with mass to Use these measurements to calculate k, the spring constant. Also calculate the angular frequency its rest length. When a mass of value M is mass at calculate the kinetic energy of the spring itself. We know that the mass has a kinetic energy of w using the measured period. placed on the end of the spring, at its new Mv2. What about the spring? We will make two basic assumptions to form a model that will equilibrium rest position, the spring is stretched allow us to find the kinetic energy of the spring. F = -kx k = -slope of F(x) a distance yo from its rest length. Figure 2: A Spring with Mass AF W = = First assumption: y = 0- This end has The mass per unit length of the spring is uniform v = 0 Your measured k and w: If we assume the spring has a mass, Ms and a length L, then any section of the spring of length & would have its 1. Draw a free body diagram and then apply Newton's First Law. Solve this equation to mass given by: Piece of find the value of yo in terms of M, k and g. spring of I- dy FBD: thickness dy 2 = Ms (or) at location yo = Comparing the Data to What You Predicted: m = eMs (9) This end 2. Adding Up the Forces - a New Variable. also moves y = L- at Um Prediction #1: w = max Xmax We know that when the mass is on the end of the spring the net force acting on the mass is given by: Prediction #2 W = "ma Vmax 5. Expressing the Total Mechanical Energy of a Mass on a Spring. EF = FS + Fo (2) k The total mechanical energy of a mass on a spring is the sum of all potential and kinetic energies Prediction number #3: W = (3) VM +MS EF = -ky + Mg for this system. Use your answers to the previous work to write an expression for the total mechanical energy of the system. (Don't forget to include the kinetic energy of the mass itself!) Note that y could have any value and the above expression (3) is the combined effect of both Making the Comparisons: gravity and the spring on the mass. We are going to define a new variable such that: E = Use the actual measured maximum values to calculate the predicted ratios: 6. Trying out a Function and Making Some Predictions. x= y - yo (4) (or) Predicted: w = _max = w = max = y = x+ yo (5) We notice from watching a mass oscillate on the end of a spring that the mass moves like a sine max Vmax This new variable is how far the mass on the end of the spring is moved from its rest or a cosine curve. Try it out! Assuming equilibrium position. Substitute equation (5) into equation (3) and also substitute in the value for yo that you found earlier. Simplify this equation. We can compare this to the more basic model taught to you in your physics class where we x(t) = Xmaxsin(wt) Note that this function can have a maximum value of Xmax assume that the spring is massless. For this simpler model: EF Find expressions for v(t) and a(t) and the maximum values these functions can have. This is quite a simple expression! To summarize this equation: x is the distance the Your Model's Prediction: w = k spring is moved from its equilibrium position (not rest length of spring!), and EF is the VM+MS combined forces of gravity and the spring added together. K_ v(t ) = The maximum value v can have is Vmax =. Simpler Model's Prediction: Write down the measured w from the previous page: w = a(t) = The maximum value a can have is amax = Is the measured value of the angular frequency closer to the prediction made by your model or to Now write out the ratio of the maximum velocity to the maximum position and the ratio the prediction made by the simpler model? of the maximum acceleration to the maximum velocity of the motion. These ratios should be in terms of the constant used in your expressions of x, v, and a Umax *max amax Umax Now substitute in your equations for x (t) and for v(t) into your answer to problem 5. Note that if friction equals zero, the total energy is supposed to be constant. The only way for this to happen is if the constants in front of the sin term are equal to the constants in front of the cos' term (since we know sin2 0 + cos2 0 = 1). Use this information to determine what value w must have, in terms of the constants of the problem, M, Ms, and k
