1. SIMPLE HARMONIC MOTION Simple hamonic motion is any motion that is equivalent to a single component of unifom cicula motion. In this situation the velocity is always geatest in the middle of the motion, and the acceleation is geatest at the edges of the motion. In simple hamonic motion, acceleation is always popotional to the negative of displacement. That is a x We can deive an equation fo simple hamonic motion by obseving a otating peg. The x-component of the position is simply the adius of the cicle, which we will call A, times the cosine of the angle, θ. The angle θ is just the poduct of the angula velocity ω and time t. Thus, the equation fo position as a function of time is x = A cosωt In many situations simple hamonic motion takes place without the involvement of any cicula motion. Fo example, a mass bouncing on a sping executes simple hamonic motion. In these cases, the vaiable A epesents the maximum displacement of the object which is known as the Amplitude. The time it takes to complete one full cycle of the motion is called the peiod, and is witten as T. We find the peiod by inceasing the angleθ =ωt by 2π and solving fo the time. Thus we find T = 2π ω The fequency, denoted by f, is the numbe of complete cycles in a given time inteval, and is equal to one ove the peiod: f = 1 T = ω 2π Equations fo velocity and acceleation can be found eithe diectly, by taking the x-component of the velocity and acceleation in unifom cicula motion, o by following
the ules of calculus to calculate the velocity and acceleation as deivatives of position. Eithe way these equations ae v = ωasinωt a = ω 2 Acosωt In geneal the object in simple hamonic motion may not begin this motion with its maximum displacement at time t=0. Including a phase facto φ povides fo this, esulting in the equations ( ) ( ) ( ) x = Acos ωt + φ v = ωasin ωt + φ a = ω 2 Acos ωt + φ The vaiables x, v, and a ae gaphed below to show how thei values compae to one anothe at diffeent points in the motion. x v a T 2T 3T Notice that, as mentioned befoe, the acceleation is popotional to the displacement, and the constant of popotionality is -ω 2. In othe wods a = ω 2 x 2
The most common use of this equation is to find the value of ω. We simply compae the equation with the equation fo acceleation that we find using the foce laws, and ead off the value. 2. MASS ON A SPRING A common example of simple hamonic motion is a mass on a sping. The foce poduced by a sping is govened by Hooke s Law, which states F = kx Hee x is the displacement of the sping fom equilibium, the point whee the sping exets no foce, and k is a constant that indicates how stiff the sping is. We can deduce fom this equation that the moe a sping is stetched, the geate the foce it exets. The negative sign indicates the sping foce is a estoing foce, that is, when an object subjected to a sping foce is displaced fom an equilibium position, the foce pushes the object towads that position. The constant k indicates how difficult the sping is to stetch o compess. Fo stiff spings k is lage, fo soft spings k is small. We can use Hooke s Law and Newton s second law ( F = ma) to study simple hamonic motion both conceptually and quantitatively. Fist we will conside the conceptual aspects of the motion. Imagine that the sping attached to a mass m lies hoizontally on a fictionless table, so that the foce of gavity does not affect the motion mass. Thus, the foce of the sping is the only foce that acts on the mass. Moving the mass to the left, esults in a lage estoing foce to the ight. Moving the mass to the ight, esults in a lage estoing foce diected to the left. Suppose we move the mass to the left and elease it. Initially, the estoing foce acceleates the mass to the ight, but as the mass begins to move, the displacement and thus the foce decease. By the time the mass eaches the equilibium position, its velocity is at a maximum, but the foce - and thus the acceleation - is 0. At this point the inetia of the mass caies it though the equilibium point to the ight. Now the sping exets a estoing foce to the left, which slows the mass. Eventually the mass stops, but by the time it does, it is well to the ight of the equilibium point and it is, once again, subjected to a lage estoing foce to the left. The pocess epeats itself, and the object continues to move back and foth, past its equilibium position. Next we conside the quantitative aspects of Hooke s law and Newton s second law. By combining the two, we obtain a = k m x We can use this to find equations fo angula velocity, peiod, and fequency fo a mass on a sping: 3
ω = T = 2π f = 1 2π k m m k k m If the sping is vetical and the mass is hanging fom it, the desciption of the motion is somewhat moe complicated as the foce of gavity is a facto. In this case the foce equation becomes kx mg = ma * This equation can be simplified to make calculations moe staightfowad. Fist conside the position, x 0, whee the mass is in equilibium (i.e. it is not acceleating). Since a=0 at this point, the equation educes to kx 0 = mg We can now eplace mg in the equation (*) above with kx 0 to find F net = k (x x 0 ) = ma The fomula is identical to the one fo a hoizontal sping on a fictionless suface, except that we have shifted ou 0 position fom the point whee the foce exeted by the sping is 0 to an equilibium point whee the sping and gavitational foces cancel. Sometimes it is useful to know the potential enegy stoed in a sping. As deived ealie the potential enegy of a sping is U sping = 1 2 kx2 The total enegy of the mass on the sping system is just the kinetic enegy plus the potential enegy: E tot = K +U sping = 1 2 mv 2 + 1 2 kx 2 = 1 2 ka 2 We obtained this equation by inputting the values v and x have when the mass is at its maximum displacement (x = A, v = 0). 4
3. PENDULUM KEY CONCEPTS Anothe system with motion that is vey nealy simple hamonic is a simple pendulum. A pendulum is just a mass at the end of a sting. A elatively simple deivation yields an equation fo the estoing foce: F = mg sinθ θ is the angle the sting makes with the hoizontal. You may ecognize the fact that mg sinθ is the component of the weight of the pendulum, mg, that is tangent to the ac descibed by the pendulum. Notice that the foce is not popotional to the angle of displacement, θ, as equied fo simple hamonic motion, but instead to the sine of θ. Thus, a pendulum is not a tue simple hamonic oscillato, and fo lage angles a pendulum deviates significantly fom simple hamonic motion. Howeve, fo small angles, less than about 15, o.26 adians, the diffeence between θ expessed in adians and sinθ is less than 1%, θ sinθ and thus a pendulum s motion though such angles is vey nealy simple hamonic motion, and we can wite: F mgθ = mg L x whee x is the displacement of the pendulum mass and L is the length of the pendulum s sting. If we now use Newton s second law we find a = g L x Fom which we can find the angula fequency (ω), the peiod (T ), and fequency (f ): ω = T = 2π f = 1 2π g L L g g L 5
4. NEWTON S LAW OF GRAVITATION KEY CONCEPTS Newton postulated that a foce exists between any two paticles in the univese and that the foce follows the elationship: F = Gm 1 m 2 That is, the foce is popotional to each of the masses, and invesely popotional to the distance between them. If eithe mass inceases the foce inceases. If the distance inceases, the foce deceases. The negative sign indicates that the foce is attactive and the unit vecto ^ (ead as hat ) shows that the foce is along the line between the two masses. Both of these ae fequently dopped fom the equation when the diection and vecto natue of the foce ae clealy undestood.) This equation is known as Newton s Law of Gavitation. Newton used the law to deive the motion of the moon and planets and then showed that his deivations ageed with ealie measuements made by othes. Newton s Law of Gavitation descibes a foce that acts at a distance, and is independent of the medium between the objects. The constant G, known as the univesal gavitational constant, was found expeimentally by Heny Cavendish by measuing small gavitational foces that exist between lage metal sphees. G has the value G = 6.67 10 11 m 3 kg s 2 = 6.67 10 11 N m 2 kg 2 Newton s law applies to individual point masses. Fo lage extended objects, it still applies if we teat an object like all of its mass is concentated at its cente of mass. Thus, we can apply the law to planets o galaxies, o even to people by just teating each object concened as though its entie mass is concentated at a single point nea thei cente. We can also use the concept of fields to incease ou undestanding of gavity. Conside a single paticle in space. Without anothe paticle, no gavitational foce exists, but we know that if anothe paticle wee placed some distance fom the fist, then a foce would be felt by the second paticle. We define the gavitational field g as the foce divided by the mass expeiencing the foce ^ g = F m It is clea fom Newton s Second Law that g is the acceleation of the paticle, and in fact that the gavitational field is simply the acceleation due to gavity. Notice that if we inset Newton s Law of gavity fo F we get: 6
g = Gm 1m 2 ^ m 2 = Gm 1 ^ Since the mass m 2 cancels, the gavitational field at a point at a distance is independent of the mass of the object subjected to the gavitational foce at that point. This is equivalent to Galileo s statement that all feely falling objects expeience the same acceleation egadless of mass. Because of this independence of mass, the gavitational field can exist without the pesence of a second mass. We can also define a potential enegy due to the gavitational field. We begin by defining gavitational potential enegy as being 0 when two objects ae infinitely fa apat. (This may seem odd, but it is in fact the most convenient convention to use fo zeo potential enegy.) With this convention we find that the potential enegy is U g = Gm 1m 2 Notice that gavitational potential enegy is negative. Because paticles have less enegy when they ae close togethe than when they fa apat, the potential enegy must be less than 0 (the value when they ae infinitely fa apat). 5. ORBITS OF PLANETS AND SATELLITES People have studied the motion of the planets fo thousands of yeas. Newton s Law of Gavitation is a poweful tool fo descibing this motion. Conside a planet moving in a cicula obit aound the sun: gavity povides the centipetal foce and we can set the equations fo the foces equal to find equations fo the centipetal acceleation, velocity, and peiod of a planet. GM s M p = M p a c a c = GM s a c = v 2 v = GM s v = 2π T T 2 = 4π2 GM 3 s 7
Notice that the centipetal acceleation, velocity, and peiod of the planet ae all independent of the planet s mass. Thus, if a tiny satellite wee at the same distance fom the sun as the eath, it would follow the same obit as the eath, at the same speed, and in the same time. The equations above can also be used fo moons o satellites obiting the eath o othe planets by eplacing the mass of the sun with the mass of the planet being obited. The enegy of an object in obit can also be detemined fom what we have leaned. Let m be the mass of the satellite, and M be the mass of the sun o planet being obited. We can wite the total enegy (kinetic plus potential) as E = K +U g = 1 2 mv 2 GMm To simplify this fo a cicula obit, we set the gavitational foce equal to the centipetal foce, as above. Altenatively, we may simply substitute in ou equation we detemined above fo v. Using eithe appoach yields this equation fo the total enegy: E = GMm 2 This is just half the potential enegy at this point. (The kinetic enegy is positive and half as big as the potential enegy so in effect the two patially cancel.) The enegy of the obiting body is negative because the satellite is bound to the sun o to the planet it is obiting. That is it has less enegy than if the two objects wee infinitely fa apat and at est. As a esult, the satellite can neve escape to infinity. Suppose we want to detemine how much enegy is equied fo a ocket to escape fom a planet. Once the ocket is infinitely fa away, its potential enegy will be 0. At that point it will have escaped, and once it has escaped, its velocity may be 0. Theefoe, the ockets total enegy must be at least 0 to escape. (If the total enegy is negative the ocket is still bound to the planet.) Setting this equal to the initial enegy we find E = K +U g = 1 mv 2 2 i GMm = 0 i K = U g v = 2GM i The subscipt i indicates the initial values of v and. Fo a ocket escaping fom the eath, the initial adius is just the adius of the eath, i =R e. Notice that it takes twice as much kinetic enegy to escape fom an obit with a given adius than to obit the planet 8
at that adius. Also the escape velocity fom an obit of a given adius is equal to the squae oot of 2 times the obital velocity at that adius. K esc = 2K ob v esc = 2 v ob 9