Simple Harmonic Motion AP Physics C

A brief refresher 𝐹=−𝑘𝑥 Hooke’s law is a restoring force, meaning that it is always trying to bring an object back to equilibrium. The net force on the object is always pointing towards equilibrium. Because of this, mass-spring systems oscillate back and forth, converting kinetic energy to potential, and vice versa. 𝐹=−𝑘𝑥

Simple Harmonic Motion Simple harmonic motion is motion that is modeled by a sinusoidal function. The amplitude is the object’s maximum displacement from equilibrium. The period is the amount of time it takes the wave to make a full cycle. Points of maximum and minimum displacement are called crests and troughs. In this diagram, A is a crest and –A is a trough. Oscillation about an equilibrium position with a linear restoring force (spring force in this case) is always simple harmonic motion.

Frequency The period of an oscillator is the amount of time required to complete a full cycle. This is analogous to the period of circular motion, which is the amount of time required to make a full circle. Frequency is the inverse of period. That means that the frequency is the # of cycles per sec. The commonly used unit is HERTZ (Hz). 𝑓= 1 𝑇 , 𝑇= 1 𝑓

Modeling Position for SHM Consider the motion of a mass-spring system. We have spent time discussing the forces and energy related to such a system, but now we will shift our focus towards modeling the position as a function of time. We will begin by analyzing the forces on the block using Newton’s second law. 𝐹 𝑛𝑒𝑡 =𝑚𝑎 −𝑘𝑥=𝑚𝑎 −𝑘𝑥=𝑚 𝑑 2 𝑥 𝑑 𝑡 2 −𝑘 𝑥 𝑚 = 𝑑 2 𝑥 𝑑 𝑡 2

Mass-Spring Systems We now have what is called a second order linear differential equation. There are different ways to find solutions to such an equation, but the best one here is what is called the “guess and check” method. I know, sounds legit. −𝑥 𝑘 𝑚 = 𝑑 2 𝑥 𝑑 𝑡 2 →𝜔= 𝑘 𝑚 −𝑥 𝜔 2 = 𝑑 2 𝑥 𝑑 𝑡 2 We define a new variable for convenience. We will see in just a second why we chose omega. What function can I take two derivatives of and get the original function, multiplied by a negative squared constant? A cosine function! Lets check and see if we are right.

Mass-Spring Systems 𝑥 𝑡 =𝐴𝑐𝑜𝑠 𝜔𝑡+𝜑 𝑑𝑥 𝑑𝑡 =𝑣 𝑡 =−𝐴𝜔 sin 𝜔𝑡+𝜑 𝑑 2 𝑥 𝑑 𝑡 2 =𝑎 𝑡 =−𝐴 𝜔 2 cos 𝜔𝑡+𝜑 𝑑 2 𝑥 𝑑 𝑡 2 =− 𝜔 2 𝐴𝑐𝑜𝑠 𝜔𝑡+𝜑 → 𝑑 2 𝑥 𝑑 𝑡 2 =− 𝜔 2 𝑥 We were right! In addition, to determining the position function, we also have functions for the velocity and acceleration. The letter omega in our function looks familiar, and that’s because it is! Omega represents the angular frequency, or angular velocity of the spring, so it can be related to our ideas of circular motion. Phi is our phase shift, it represents the initial conditions of our object.

Period of a Spring Since all springs exhibit properties of circle motion we can use these expressions to derive the formula for the period of a spring. We will use our angular frequency from our position function to derive the period of a spring. 𝑣=𝑟𝜔 𝑣= 2𝜋𝑟 𝑇 𝜔= 𝑘 𝑚 𝑟𝜔= 2𝜋𝑟 𝑇 →𝑇= 2𝜋 𝜔 → 𝑇 𝑠𝑝𝑟𝑖𝑛𝑔 =2𝜋 𝑚 𝑘

Example A body oscillates with the simple harmonic motion according to the equation: 𝑥 𝑡 =6cos⁡(3𝜋𝑡+ 𝜋 3 ) Calculate the position, the velocity, and the acceleration, at the time t = 2.0 s. Also, find the frequency and the period of motion.

Simple Pendulums We can use a very similar method as we did before to determine a position function and the period of a simple pendulum. A simple pendulum is one where the mass in concentrated at the end of the pendulum. We will begin by examining the forces acting on the pendulum when it is displaced from equilibrium. The tangential component of gravity is the restoring force here, and because it is restoring we define it as being negative.

Differential Equations We will use Newton’s second law for rotation to derive a angular position function. 𝜏 𝑛𝑒𝑡 =𝐼𝛼 −𝑚𝑔𝑙𝑠𝑖𝑛 𝜃 =𝑚 𝑙 2 𝑑 2 𝜃 𝑑 𝑡 2 −𝑔𝑠𝑖𝑛 𝜃 =𝑙 𝑑 2 𝜃 𝑑 𝑡 2 − 𝑔 𝑙 𝑠𝑖𝑛 𝜃 = 𝑑 2 𝜃 𝑑 𝑡 2 There is a problem here. The differential equation is second order, but the sine term makes it non-linear. Solutions to non-linear differential equations require initial values and direct computation. We will confine this scenario to small angles, <20 degrees, so that we can use the small angle approximation. 𝐼𝑓 𝜃≪1, 𝑡ℎ𝑒𝑛 sin 𝜃 ≈𝜃

Differential Equations This linearizes our differential equation, making it solvable using our guess and check method. − 𝑔 𝑙 𝜃= 𝑑 2 𝜃 𝑑 𝑡 2 →ω= 𝑔 𝑙 − 𝜔 2 𝜃= 𝑑 2 𝜃 𝑑 𝑡 2 This gives us the same solution that we had with an oscillating mass-spring system 𝜃 𝑡 =𝐴𝑐𝑜𝑠 𝜔𝑡+𝜑 We can also use our relationship between angular frequency and period to find the period of a pendulum for small oscillations. 𝑇= 2𝜋 𝜔 →𝜔= 𝑔 𝑙 𝑇 𝑝𝑒𝑛𝑑𝑢𝑙𝑢𝑚 =2𝜋 𝑙 𝑔

Example Calculate the length of a simple pendulum that marks seconds by completing a full cycle every 2 s.

Physical Pendulums A physical pendulum is an oscillating body that rotates according to the location of its center of mass rather than a simple pendulum where all the mass is located at the end of a light string. A rod or a meter stick would be examples of physical pendulums. The difference mathematically is that the gravitational torque is smaller, and the moment of inertia will be different. Generally speaking, we will define the period of a physical pendulum as: 𝑇 𝑝ℎ𝑦𝑠𝑖𝑐𝑎𝑙 𝑝𝑒𝑛𝑑𝑢𝑙𝑢𝑚 =2𝜋 𝐼 𝑚𝑔𝑑 𝐼=𝑚𝑜𝑚𝑒𝑛𝑡 𝑜𝑓 𝑖𝑛𝑒𝑟𝑡𝑖𝑎 𝑑=𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑓𝑟𝑜𝑚 𝑝𝑖𝑣𝑜𝑡 𝑡𝑜 𝑐𝑒𝑛𝑡𝑒𝑟 𝑜𝑓 𝑚𝑎𝑠𝑠 (𝑚𝑜𝑚𝑒𝑛𝑡 𝑎𝑟𝑚)