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Published byCristian Harmon Modified over 2 years ago

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Simple Harmonic Motion Sinusoidal Curve and Circular Motion

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A mass is oscillating on a spring Position in equal time intervals: This type of motion, is called Simple Harmonic Motion. The system is trying to place the mass at its static equilibrium position

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Repetitive Motion Repetitive motion can be modelled as circular motion A is the equivalent to x the displacement from equilibrium and is also the radius of the circle

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What Can Algebra Tell Us? Circle Spring This tells us that the period only depends upon the mass and the spring constant, not on the distance the spring is pulled back. In English, the time for the mass to move back and forth is ALWAYS the same even if the amplitude decreases.

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What Can Algebra Tell Us? Circle Spring This tells us that the velocity depends upon the inverse of the mass, the spring constant, and the distance the spring is pulled back. In English, the velocity of the spring depends on how far it was pulled pack and the spring constant as well as the bigger the mass the slower the velocity of the spring. We could has also come up with the equations by setting the Kinetic Energy equal to the Spring Potential Energy and solving for v

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What Can Algebra Tell Us? Circle Spring This tells us that the spring experiences no acceleration when r=0, this is, at the relaxed position. But we have to use logic to determine which way the acceleration is pointing.

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Calculus Can Tell us More Circle From Force by Spring Therefore x is a function of time, that when given a time value, t, I can tell you the position x, where the mass is. But, the question is: What function’s second derivative is equal to this value? The force by the spring is called a Restoring Force

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Circular Motion Yes, But what if I didn’t Know Calculus?

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Vertical position versus time: Period T

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Sinusoidal motion Time (s) Displacement (cm) Period T

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Sine function: mathematically x y 2π2π π/2π3π/22π2π5π/23π3π7π/24π4π9π/25π5π 1 y=sin(x) y=cos(x)

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Sine function: employed for oscillations x y π/2 π 3π/2 2π2π5π/2 3π3π 7π/24π4π 9π/2 5π5π 1 y=sin(x) Time t (s) Displacement y (m) T/2 T2T -A A y= A sin(ωt)

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Sine function: employed for oscillations 1. Maximum displacement A 2. ωT = 2π 3. Initial condition y(t=0) Angular frequency in rad/s Amplitude A is the maximum distance from equilibrium Starting from equilibrium: y=A sin(ωt) Starting from A: y=A cos(ωt) Starting from -A: y=-A cos(ωt) Therefore Note:

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Position, Velocity, Acceleration

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Example 1 - determine y(t) y(cm) t(s) Period? T=4 s Sine/cosine? Sine Amplitude? 15 cm Where is the mass after 12 seconds?

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Example 2 – graph y(t) Amplitude? 3cm -3 3 y (cm) y(t=0)? -3cm Period? 2s t(s) When will the mass be at +3cm? 1s, 3s, 5s, … When will the mass be at 0? 0.5s, 1.5s, 2.5s, 3.5 s …

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Summary Harmonic oscillations are sinusoidal Motion is repeated with a period T Motion occurs between a positive and negative maximum value, named Amplitude Position: y=A sin(ωt) or y=A cos(ωt) Velocity: v=Aωcos(ωt) or v=-Aωsin(ωt) Acceleration: a=-Aω 2 sin(ωt) or a=-Aω 2 cos(ωt) Angular frequency:

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Energy of Harmonic Motion According to the law of conservation of energy, when the mass of attached to a spring is released, the total energy of the system remains constant. This energy is the sum of the elastic potential energy and the kinetic energy of the spring

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Energy Example A 55 g box is attached to a horizontal spring (force constant 24 N/m). The spring is then compressed to a position A= 8.6 cm to the left of the equilibrium position. The box is released and undergoes SHM. a)What is the speed of the box when it is at x=5.1 cm from the equilibrium position? b)What is the maximum speed of the box? Can you solve these questions two ways? Using energy? Using position and velocity formulas?

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Energy Example A 55 g box is attached to a horizontal spring (force constant 24 N/m). The spring is then compressed to a position A= 8.6 cm to the left of the equilibrium position. The box is released and undergoes SHM. a)What is the speed of the box when it is at x=5.1 cm from the equilibrium position?

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Energy Example Let the position function be: Let the velocity function be: Set the position function = to 5.1 and solve for the t the time. Plug this t value into the velocity function Therefore the speed is 1.4 m/s

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A 55 g box is attached to a horizontal spring (force constant 24 N/m). The spring is then compressed to a position A= 8.6 cm to the left of the equilibrium position. The box is released and undergoes SHM. b) What is the maximum speed of the box? Energy Example The maximum speed will occur when it is at the equilibrium position (x f =0)

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b) What is the maximum speed of the box? Energy Example The maximum speed will occur when it is at the equilibrium position (x f =0) Let the position function be: Let the velocity function be: Set the position function = to 0 and solve for the t the time. Plug this t value into the velocity function Therefore the speed is 1.8 m/s

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Conservation of Energy When the spring is in Equilibrium with gravity, the forces are balanced (the downward force of gravity matches the upward for of the spring. So: Where ‘d’ is the equilibrium extension of the spring and ‘g’ is the acceleration due to gravity. Therefore : d

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Conservation of Energy We will now set the spring bouncing. At any time the total energy is: To simplify things. I’m going to choose U g = 0 at the extension equilibrium. I’m free to choose any location for this. This gives us: x=0 +x d U g =0

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Conservation of Energy Now, let’s expand out this equation:

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Conservation of Energy Now, let’s expand out this equation: I’m a constant Therefore, I’m a constant too!

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Java Site for SHM shm/springEnergy/simulate/page2.ht ml

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