Simple Harmonic Motion Ch. 11 Section 1

Felastic = kDx Hooke’s Law Felastic is the force restoring the spring to the equilibrium position. A minus sign is needed because force (F) and displacement (x) are in opposite directions. k is the spring constant in N/m. k measures the strength of the spring.

Simple Harmonic Motion file:///F:/Honors%20Physics/Topics/T8-Vibrations%20and%20Waves/videos/70350.html

Classroom Practice Problem A slingshot consists of two rubber bands that approximate a spring. The equivalent spring constant for the two rubber bands combined is 1.25  103 N/m. How much force is exerted on a ball bearing in the leather cup if the rubber bands are stretched a distance of 2.50 cm? Answer: 31.2 N For problems, it is a good idea to go through the steps on the overhead projector or board so students can see the process instead of just seeing the solution. Allow students some time to work on problems and then show them the proper solutions. Do not rush through the solutions. Discuss the importance of units at every step. Problem solving is a developed skill and good examples are very helpful. Use F = -kx and make sure students are careful about the units (cm and m).

Simple Harmonic Motion Simple harmonic motion (SHM) results from systems that obey Hooke’s law. SHM is a back and forth motion that obeys certain rules for velocity and acceleration based on F = -kx. The next slide works through a detailed example of SHM for a mass-spring system.

Spring Constant Click below to watch the Visual Concept. file:///F:/Honors%20Physics/Topics/T8-Vibrations%20and%20Waves/videos/70641.html file:///F:/Honors%20Physics/Topics/T8-Vibrations%20and%20Waves/videos/70641.html

Simple Harmonic Motion Where is the force maximum? a and c Where is the force zero? b Where is the acceleration maximum? Where is the acceleration zero? Where is the velocity maximum? Where is the velocity zero? Lead students through these answers. Since a = F/m, the acceleration follows the same pattern as the force. The velocity reaches a maximum at the equilibrium point because the force (and acceleration) is just switching directions at that point.

The Simple Pendulum The pendulum shown has a restoring force Fg,x. A component of the force of gravity At small angles, Fg,x is proportional to the displacement (), so the pendulum obeys Hooke’s law. Simple harmonic motion occurs. The next slide reinforces the dependence on small angles by having students calculate displacements for small and large angles.

The Simple Pendulum Find the restoring force at 3.00°, 9.00°, 27.0°, and 81.0° if Fg = 10.0 N. Are the forces proportional to the displacements? Have students make the calculations before showing the video on the next slide. The restoring force is simply the x-component of the force of gravity (Fg sin ). When tripling the displacement from 3° to 9° to 27° to 81°, the restoring force triples, then almost triples, then just barely doubles. So, the restoring force is proportional to the displacement for small angles (generally <15°). Thus, the pendulum follows SHM as long as the angles are small. For large angles, the motion is more complex.

Restoring Force  3.00o 0.523 N 3x x3 9.00o 1.56 N 3x x2.9 27.0o 4.54 N 3x x2.2 81.0o 9.88 N

The Simple Pendulum Are the restoring forces proportional to the displacements? Answer: only for small angles Generally, the angle must be ≤ 15o Have students make the calculations before showing the video on the next slide. The restoring force is simply the x-component of the force of gravity (Fg sin ). When tripling the displacement from 3° to 9° to 27° to 81°, the restoring force triples, then almost triples, then just barely doubles. So, the restoring force is proportional to the displacement for small angles (generally <15°). Thus, the pendulum follows SHM as long as the angles are small. For large angles, the motion is more complex.

Restoring Force and Simple Pendulums Click below to watch the Visual Concept. Visual Concept

Measuring Simple Harmonic Motion Ch. 11. 2

Measuring Simple Harmonic Motion Amplitude (A) is the maximum displacement from equilibrium. SI unit: meters (m) or radians (rad) Period (T) is the time for one complete cycle. SI unit: seconds (s) Frequency (f) is the number of cycles in a unit of time. SI unit: cycles per second (cycles/s) or s-1 or Hertz (Hz) It is important to stress the difference between frequency and period. One is seconds/cycle and the other is cycles per second, so each is the inverse of the other. In order to keep units straight when using frequency, the easiest unit is s-1. When giving students frequencies, instead of saying the frequency is 20 Hz or 20 cycles per second, it is useful to say the frequency is 20 per second. This will get them accustomed to the terminology. When entering the value into equations, they should enter it as 20/second. This facilitates the problem solving aspect. It should eventually become automatic for them to equate Hz with “per second” and vice versa.

Relationship of T and f

Period of a Simple Pendulum Simple pendulums small angles (<15°) The period (T) depends only on the length (L) and the value for ag. Mass does not affect the period. All masses accelerate at the same rate.

Period of a Mass-Spring System Greater spring constants  shorter periods Stiffer springs provide greater force (Felastic = -kx) and therefore greater accelerations. Greater masses  longer periods Large masses accelerate more slowly. Point out to students that these are square root relationships. Quadrupling the mass provides a period that is twice as large.

Classroom Practice Problems What is the period of a 3.98-m-long pendulum? What is the period and frequency of a 99.4-cm-long pendulum? Answers: 4.00 s, 2.00 s, and 0.500 s-1 (0.500/s or 0.500 Hz) A desktop toy pendulum swings back and forth once every 1.0 s. How long is this pendulum? Answer: 0.25 m For problems, it is a good idea to go through the steps on the overhead projector or board so students can see the process instead of just seeing the solution. Allow students some time to work on problems and then show them the proper solutions. Do not rush through the solutions. Discuss the importance of units at every step. Problem solving is a developed skill and good examples are very helpful. The 2nd part of the first problem asks for frequency as well as period (f = 1/T). Notice on these problems that the period keeps cutting in half (4 --> 2 --> 1) while the lengths are reduced by one-fourth (3.98 ---> 0.994--->0.25). When the unknown is under the radical sign, an easy approach is to simply square both sides of the equation : T2 = 42 (L/g).

Classroom Practice Problems What is the free-fall acceleration at a location where a 6.00-m-long pendulum swings exactly 100 cycles in 492 s? Answer: 9.79 m/s2 A 1.0 kg mass attached to one end of a spring completes one oscillation every 2.0 s. Find the spring constant. Answer: 9.9 N/m For problems, it is a good idea to go through the steps on the overhead projector or board so students can see the process instead of just seeing the solution. Allow students some time to work on problems and then show them the proper solutions. Do not rush through the solutions. Discuss the importance of units at every step. Problem solving is a developed skill and good examples are very helpful.