Lesson 44: Acceleration, Velocity, and Period in SHM

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1 Lesson 44: Acceleration, Velocity, and Period in SHM Since there is a restoring force acting on objects in SHM it akes sense that the object will accelerate. In Physics 20 you are only required to explain this acceleration for asses on horizontal springs with no friction, and basic pendulus. Vertical springs will not be covered here. Acceleration of a Mass on a Spring As a ass bounces back and forth on a spring, it will have a changing acceleration. he changing acceleration happens because the restoring force is always changing. As long as the situation is frictionless, there are no other forces to consider, so the net force will be the restoring force. F NE =F s a= kx a= kx Exaple 1: Deterine the acceleration of a 0.250kg ass on the end of a 54.9N/ spring if it has been... a) stretched 12 c fro its equilibriu and released. b) copressed 18 c fro its equilibriu and released. a) a= kx a= 54.9(0.12) a= = 26/s 2 b) a= kx a= 54.9( 0.18) a=39.528=40 / s 2 hese answers are the acceleration at that oent. An instant later (when they have oved to a different position) their accelerations will be different. he answer in (a) is negative because the spring has been stretched; the ass is trying to accelerate in the opposite direction back to equilibriu. In (b) the answer is positive because the object was copressed and the ass is trying to accelerate back in the positive direction to equilibriu. acceleration acceleration Illustration 1: Mass undergoing acceleration. 1/24/2016 studyphysics.ca Page 1 of 5 / Section 7.3

2 Velocity of a Mass on a Spring Soe people think that a big acceleration autoatically eans that the object is oving at a big velocity. Reeber that in the exaples above the ass has just been released fro rest, so its velocity starts at zero. As it accelerates back to the equilibriu point, the elastic potential energy that was stored in the spring is converted to kinetic energy as the ass goes faster and faster. At the equilibriu point there is no elastic potential energy reaining, so it has all becoe kinetic energy (this is still assuing a frictionless surface). It is at the equilibriu point that the ass is oving at its axiu velocity. We will call the furthest that the ass is fro the equilibriu point the aplitude (A) of the spring. It is still easured in etres. If we want to calculate the axiu velocity of a ass on a spring (as it passes through its equilibriu point) when it is released with a certain aplitude, we can use conservation of energy... E =E ' E k E p =E k ' E p ' E p =E k ' 1 2 kx 2 ax= 1 2 v 2 ax 1 2 ka2 = 1 2 v 2 ax ka 2 2 = 2 = ka2 ka2 = =A k = axiu velocity at equilibriu (/s) A = aplitude of ass () k = spring constant (N/) = ass (kg) Exaple 2: A 17kg ass is pulled 13c away fro its equilibriu point, on a spring with a 367 N/ constant. Deterine its axiu velocity as it passes through equilibriu. 1/24/2016 studyphysics.ca Page 2 of 5 / Section 7.3

3 = A k = = =0.60 /s Period of a Mass on a Spring It is possible to show that SHM very closely resebles circular otion. We won't be going in to all the reasons here, but I' hoping you will trust e. he one thing we need to keep in ind for the following anipulations to ake sense is that we can consider the radius of a circle (r) to be the sae as the axiu aplitude (A) of an object in SHM. his eans we can play around with an old forula fro circular otion this way... v= 2 r = 2 A We can ake this derived forula equal to the forula fro the last section... =v A ax k = 2 A k 2 = = 2 k =2 k = period (s) = ass (kg) k = spring constant (N/) Exaple 3: Using the inforation fro the previous exaple, deterine the period of the ass. 1/24/2016 studyphysics.ca Page 3 of 5 / Section 7.3

4 =2π k =2(3.14) = =1.4 s Warning! I used 3.14 for pi, which gave e three sig digs. Do not use the pi button on your calculator. Period of Pendulus One of the first people to realize how physics could explain pendulus was Galileo. According to Galileo's own notes, he was visiting a local church one day when he happened to look at a person lighting the laps that were hanging in the church. hese lanterns were hanging fro the roof by long chains. he lanterns theselves were quite heavy. he person lighting the lanterns was using a very long pole with a taper (fancy thin candle) on the end to light the each tie he reached up with the pole, he would nudge the lantern a bit which started it swinging. Galileo noticed that although all the lanterns were the sae ass, the length of the chain they were dangling fro seeed to change how fast they were swinging. In order to get soe sort of easureent of the tie it took the to swing, he actually used his own pulse to tie the! Although we have coe a long way fro the work that Galileo started, we still have a pretty siple equation that we can use to figure out how long it takes for a pendulu to swing =2 l g = period of one swing (s) l = length of wire () g = gravity (/s 2 ) Notice that the ass of the bob (the weight at the end of a pendulu) is not in the forula. his agrees with Galileo s observation that different asses on the end of the pendulu will not affect the period of the swing. he only thing a heavy bob does is keep the pendulu going for a longer period of tie before stopping. hink of it ters of inertia a heavy bob has lots of inertia, so the pendulu swings back and forth for a long tie before it stops. A lighter bob has little inertia, so it only swings for a little while before coing to rest at its equilibriu position. What really atters in the case of the pendulu is the length of the pendulu. If you ve ever looked carefully at the pendulu on a Grandfather clock, you ight have noticed that the bob can ove up and down a little. his is to adjust the period of the pendulu so that the clock runs at the right speed. Be careful when you are using this forula. Reeber three things: 1. he period is the tie it takes to coplete one full swing that eans if you let go of the bob (the weight on the end), it will swing away fro your hand and back to your hand. hat s one 1/24/2016 studyphysics.ca Page 4 of 5 / Section 7.3

5 coplete swing. Back to where it started and ready to ove in the original direction again. 2. his forula only works well if the pendulu is held at an angle of less than 15 fro equilibriu position. As the angle gets further past 15 errors start to creep in. 3. Be careful if you need to solve it for l or g. You ll see in the following exaples how to do this. Exaple 4: Deterine the period of a pendulu that is 12.5 long. Since we have no reason to think otherwise, we will assue that it is on Earth. =2(3.14) = =7.09 s Exaple 5: We decide to easure gravity in a particular location on Earth. I use a 2.75 long pendulu and find that it has a period of 3.33 s. Deterine the acceleration due to gravity in this area. g= 4π 2 l 2 g= 4(3.14)2 (2.75) g= =9.78/s 2 Exaple 6: Deterine the length of pendulu that would have a period of one inute. We can already ake a pretty good guess that this is going to be a pretty long pendulu to take that long to swing. Hoework 372 # #1, #1-3 l= g 2 4π 2 l= 9.81(60)2 4(3.14) 2 l= =895 Yikes! Alost one kiloetre long. And before you get grupy about sig digs, notice that in the question I said one inute, which is by definition exactly 60 seconds, so I had an infinite nuber of sig digs there. Only the easureent of the acceleration due to gravity and pi have sig digs I need to worry about (three on each). 1/24/2016 studyphysics.ca Page 5 of 5 / Section 7.3