Oscillation - Unit 6

Oscillation - Unit 6 Simple harmonic motion, or oscillation, can be described with the behaviour of a mass on a spring. If you put a mass on a spring and then compress or stretch it a little, it will expand or contract back into its equilibrium position: however, it will have some velocity because of that. It will then continue its motion until the force of the spring fully stops it. However, at that point, the spring would be slightly compressed or stretched again. It will keep on oscillating forever in ideal situations. The relationship for the position of the mass in SHM is given by the following equation: 𝑥 (𝑡) = 𝐴 𝑐𝑜𝑠 𝑐𝑜𝑠 (𝜔𝑡 − 𝜙) A is the amplitude of the SHM, and 𝜔 is the angular frequency of the SHM, while 𝜙 is the phase shift. The relationships for velocity and acceleration are: 𝑣(𝑡) = −𝐴𝜔 𝑠𝑖𝑛 𝑠𝑖𝑛 (𝜔𝑡 − 𝜙) 𝑎(𝑡) = −𝐴𝜔2 𝑐𝑜𝑠 𝑐𝑜𝑠 (𝜔𝑡 − 𝜙) = −𝜔2 𝑥 Now, why is this? We’ll need to think with some calculus. You won’t need to fully understand the exact calculus as it’s a little beyond the scope of the course, but try to follow the general reasoning. First, we start with Hooke’s Law. 𝐹𝑠 = −𝑘𝑥 We also know that 𝐹 = 𝑚𝑎 And that 𝑑2 𝑥 𝑎= 2 𝑑𝑡 We can now equate them. 𝑑2 𝑥 −𝑘𝑥 = 𝑚 2 𝑑𝑡 𝑑2 𝑥 𝑘 + 𝑥=0 𝑑𝑡 2 𝑚 This is called a second-order ordinary differential equation with constant coefficients, in standard form. To solve this ODE, we want to look for a function of x(t) where the second derivative with respect to time will look like x, but with a negative sign. The most likely candidates are the trigonometric functions sin(t) and cos(t). We start with cos(t) because cos(0) is 1, and most oscillations begin at maximum amplitude A (you pull and hold the mass on a spring before letting it go, for example). You can always alter the phase shift to get equivalent functions, so it doesn’t really matter. We start with the general form of it: 𝑥 (𝑡) = 𝐴 𝑐𝑜𝑠 𝑐𝑜𝑠 (𝜔𝑡 − 𝜙) When we take the derivatives of this, we get: 𝑣(𝑡) = −𝐴𝜔 𝑠𝑖𝑛 𝑠𝑖𝑛 (𝜔𝑡 − 𝜙) 𝑎(𝑡) = −𝐴𝜔2 𝑐𝑜𝑠 𝑐𝑜𝑠 (𝜔𝑡 − 𝜙) = −𝜔2 𝑥 We can substitute this value of a(t) back into our original equation to find the value of 𝜔. 𝑑2 𝑥 𝑘 + 𝑥=0 𝑑𝑡 2 𝑚 −𝜔2 𝑥 + 𝑘 𝑘 𝑥 = 0 ⟶ 𝜔2 𝑥 = 𝑥 𝑚 𝑚 𝑘 𝑚 𝜔=√ Other Types of SHM: the Pendulum You can do this same rigmarole with all sorts of different physical situations. Any system creating a linear restoring force will exhibit SHM characteristics. For example, you can examine a simple pendulum at small angles and find that there is essentially a linearly-related restoring force, because of the small-angle approximation. The angular frequency is described in terms of properties of the pendulum (at small angles). For a simple pendulum: 𝑔 𝑙 𝜔=√ Here, l is the length of the pendulum. You can also do this for physical pendulums, which are pendulums that are objects with rotational inertia rotating on a pivot. 𝑚𝑔𝐿𝑐𝑚 𝐼 𝜔=√ Here, I is the rotational inertia of the swinging object, while Lcm is the distance from the center of mass that the pivot point is at. Note that if you pivot it through the center of mass, there cannot be oscillation. Period The period of the SHM can be found if you look at how cos(t) behaves. 𝑇= 2𝜋 1 = 𝜔 𝑓 We can think about the period for a few different types of SHM. For a mass on a spring: 𝑇 = 2𝜋√ 𝑚 𝑘 For a simple pendulum (mass on a string): 𝑇 = 2𝜋√ 𝑙 𝑔 For a physical pendulum: 𝑇 = 2𝜋√ 𝐼 𝑚𝑔𝐿𝑐𝑚 The frequency for each of these will just be the reciprocal. Energy in Oscillation The total energy in a spring-mass system is: 1 𝐸𝑡𝑜𝑡𝑎𝑙 = 𝑘𝐴2 2 It is the sum of the 2 sources of mechanical energy: 𝐸𝑡𝑜𝑡𝑎𝑙 = 𝐾 + 𝑈𝑠