Learning object harmonic motion
- 1. Physics Learning Object SIMPLE HARMONIC MOTION OF A MASS ON A VERTICAL SPRING
- 2. Consider a 2.0kg mass hanging from a vertical spring with a spring constant, k, of 200N/m. The mass is oscillating, completing 1 oscillation every 2.0 seconds. At a given time, the mass is travelling at 1.0m/s when it is 0.1m away from the equilibrium position. If we consider the distance of 0.1m as the starting point of the oscillation (at t=0), a) Write an equation to model the position of the simple harmonic motion described using the cosine function. b) Determine the intervals during which the displacement is negative. c) Determine the maximum velocity that the mass achieves.
- 3. Begin the problem by designing a sketch of the situation: Equilibrium position K=200N/m Δx=0.1m v=1.0m/s 2.0kg + It is important to remember to assign a sign convention
- 4. a) Write an equation using the cosine function to model the position of the simple harmonic motion described. Consider the equation of simple harmonic motion: x(t)=Acos(ωt+Φ) In order to write the equation for this particular situation, we must determine constants A, ω, and Φ
- 5. To find the constant A, we must first find the total energy of the system. Since the mass is already in motion, we can think about the total energy as the sum of the potential energy and of the kinetic energy: Etotal=PE+KE Substituting the work equation for a spring into the potential energy, we get: Etotal=1/2kΔx2+1/2mv2
- 6. Now plug in the given values and solve for Etotal: k=200N/m m=2.0kg Δx=0.1m v=1.0m/s Etotal=1/2kΔx2+1/2mv2 Etotal=1/2(200)(0.1)2+1/2(2.0)(1.0)2 Etotal=1+1 Etotal= 2 J
- 7. Now that we have the total energy, we can solve for the maximum amplitude. At the maximum amplitude, all of the energy in the system is transformed into potential energy: Etotal=PE+KE Etotal=1/2kΔx2 We can now solve for Δx Etotal=1/2kΔx2 2=1/2(200)Δx2 0.02=Δx2 0.14m=Δx Since the amplitude is equal to Δx, A=0.14 0
- 8. Now we need to find the angular frequency, ω. We know the time it takes for one period is 2.0s, so we can easily solve for ω: ω=2π/T ω=2π/2 ω=π
- 9. Plugging in A and ω into the equation for simple harmonic motion we get: x(t)=0.14cos(πt+Φ) We also know that at t=0, the mass is 0.1m away from the equilibrium point: x(0)=0.14cos(π(0)+Φ) Simplify: 0.1=0.14cos(Φ) 0.714=cos(Φ) cos-1(0.714)=Φ 0.775=Φ
- 10. So the equation of the function that describes the position of the mass as a function of time is: x(t)=0.14cos(πt+0.775)
- 11. b) Determine the intervals during which the displacement is negative. To answer this question we need to first find the zeros of the function and then use the symmetry of the cosine function to find where these zeros repeat. We begin by setting the cosine function to zero: 0=0.14cos(πt+0.775) Simplify: 0=cos(πt+0.775) cos-1(0)=πt+0.775 1.57-0.775=πt 0.253=t
- 12. The first zero of this function is at t=0.253 and we know from the question that it will repeat this oscillation every 2 seconds, crossing the x-axis once in between this oscillation. So the zeros of the function can be described as: 0=x(0.253) 0=x(0.253+1) 0=x(0.253+2) 0=x(0.253+2) Etc.. 0=x(0.253+n), where n is an integer
- 13. Now we plug in a point between any of these zeros to determine whether the value is positive or negative: X(0.5)=0.14cos(π(0.5)+0.775) X(0.5)= -0.098 Since the value is negative we know that the interval (0.253,1.253) is negative, and repeats every 2s. So the displacement is negative for (0.253+2n,1.253+2n), where n is an integer
- 14. This graph of the function corresponds to our solution
- 15. c) Determine the maximum velocity that the mass achieves. Again we can think about this problem in terms of energy. At maximum velocity, all of the potential energy of the system has been transformed into kinetic energy. (Note that this occurs when the object passes through the equilibrium point).
- 17. Therefore, Etotal=KE Etotal=1/2mv2 2=1/2(2)v2 2=v2 ±√2=v When you square root a number, you can have both a positive and negative value, which makes sense in this situation either as describing the velocity of the object as it moves up or down. This ties in the importance of the sign convention that we assigned at the beginning of the solution.