LECTURE 3 ENERGY AND PENDULUM MOTION. Instructor: Kazumi Tolich

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LECTURE 3 ENERGY AND PENDULUM MOTION Instructor: Kazumi Tolich

Lecture 3 2 14.4: Energy in simple harmonic motion Finding the frequency for simple harmonic motion 14.5: Pendulum motion Physical pendulum and locomotion

14.4 Energy in simple harmonic motion 3 If a system of an object with a mass m on a horizontal spring with a spring constant k is isolated, mechanical energy is conserved. E = K + U = ( ) mv) + ( ) kx) = constant.

Quiz: 14.4-1 When does a body moving in simple harmonic motion has the maximum acceleration? Choose all that apply. When it has A. maximum velocity. B. zero velocity. C. maximum kinetic energy. D. minimum kinetic energy. E. maximum potential energy. F. minimum potential energy. G. zero displacement.

Quiz: 14.4-1 answer A body moving in simple harmonic motion has the maximum acceleration When it has zero velocity. minimum kinetic energy. maximum potential energy. When the spring is maximally stretched/compressed, the force on the mass is maximum resulting in the maximum acceleration. At x = ±A, E = U 456 = ( ) ka) At x = 0, E = K 456 = ( ) m v 456 )

Quiz: 14.4-2 6 A mass oscillates in simple harmonic motion with amplitude A. If the mass is doubled, but the amplitude is not changed, what will happen to the total mechanical energy of the mass-spring system? A. Increases B. Stays the same C. Decreases

Quiz: 14.4-2 answer 7 Stays the same The total mechanical energy is equal to the maximum elastic potential energy, which is U = 8 9 ka). This does not depend on mass, so a change in mass will not affect the energy of the system. At the equilibrium position, the total mechanical energy is in its kinetic energy K = ( ) mv). So, if the mass is doubled, it must be moving slower so that the maximum kinetic energy stays the same.

Quiz: 14.4-3 A block oscillates on a very long horizontal spring. The graph shows the block s kinetic energy as a function of position. What is the spring constant in N/m?

Quiz: 14.4-3 answer From conservation of energy, E = U 456 = K 456 = ( ) ka). k = ): ;<= > 9 = )? A ) 4 9 = 4 C 4

Quiz: 14.4-4 A mass oscillates on a horizontal spring with period T = 2.0 s. If the amplitude of the oscillation is doubled, what is the new period in seconds?

Quiz: 14.4-4 answer 2.0 s Energy conservation yields the frequency and period of oscillation: K 456 = ( ) m v 456 ) = U 456 = ( ) ka) and v 456 = 2πfA (from Section 14.3) The frequency and period of simple harmonic motion are determined by the physical properties of the oscillator and do not depend on the amplitude A. f = 1 2π k m T = 2π m k

Quiz: 14.4-5 Two identical blocks oscillate on different horizontal springs. Which spring has the larger spring constant? A. The red spring B. The blue spring C. Both the same D. There s not enough information to tell.

Quiz: 14.4-5 answer The red spring has the larger spring constant. The red spring has a shorter period. T = 2π L M T T

14.5 Pendulum motion 14 The small oscillation of a pendulum is a simple harmonic motion. The arc length and the angle are given by: s t = A cos 2πft θ t = θ 456 cos 2πft The frequency and the period of oscillation are given by f = 1 2π g L T = 2π L g

Quiz: 14.5-1 A series of pendula with different length strings and different masses is shown below. Each pendulum is pulled to the side by the same (small) angle, the pendulums are released, and they begin to swing from side to side. Rank the pendula according their frequency, smallest first. 20 cm

Quiz: 14.5-1 answer / Demo C < B = D < A f = ( )X Y Z The frequency, and hence the period, is independent of the mass. Demo: Simple pendula with different lengths and masses 20 cm

Quiz: 14.5-2 17 The graph shows the square of the period versus the length of a simple pendulum on a certain planet. What is the acceleration due to gravity on that planet in m/s 2?

Quiz: 14.5-2 answer 18 4 m/s 2 T = 2π Z Y g = 2π ) Z = 2π ) \ 4 = 4 m s ) [ 9 \] ^9

Quiz: 14.5-3 19 Two pendula have identical periods. One has a slightly larger amplitude than the other, but both swing through small angles from the vertical. Which of the following must be true of the pendulum that has the larger amplitude? Choose all the apply. A. It is longer than the other one. B. It is shorter than the other one. C. It has slightly more energy than the other one. D. It has slightly less energy than the other one. E. It moves faster at the lowest point in its swing than the other one. F. It moves slower at the lowest point in its swing than the other one.

Quiz: 14.5-3 answer 20 It moves faster at the lowest point in its swing than the other one. The two pendula must be of the same length as their periods are identical. T = 2π Z Y The greater the amplitude, the greater distance and therefore faster on average that the bob must travel. The energy of the pendulum depends on the mass of the bob, but we have no information about it. The mechanical energy of the bob-earth system is conserved. As the pendulums swing down, the gravitational potential energy of the bob-earth system is converted into kinetic energy. Gravitational potential energy depends on the mass of the bob. K 456 = ( ) m v 456 ) = U 456 = mgh

14.5 physical pendulums and locomotion / Demo 21 A physical pendulum is a rigid object free to rotate about a horizontal axis that is not through its center of gravity that oscillates when displaced from equilibrium. Physical pendula do not exhibit true simple harmonic motion for any angle. However, if the angle of oscillation is small, the motion is close to and can be modeled as simple harmonic motion whose frequency is given by f = 1 2π mgd I

Quiz: 14.5-4 (Knight P14.31) 22 Interestingly, there have been studies using cadavers to determine the moment of inertia of human body parts by letting them swing as a pendulum about a joint. In one study, the center of gravity of a 5.0 kg lower leg was found to be 18 cm from the knee. When pivoted at the knee and allowed to swing, the oscillation frequency was 1.6 Hz. What is the moment of inertia of the lower leg in kgdm 2?

Quiz: 14.5-4 answer 23 8.7 10 h) kg d m ) Treat the lower leg as a physical pendulum. f = ( )X LYk I = LYk n.] op )Xm 9 = l q.?] 4 ^9 ].(? 4 = 8.7 10 h) kg d m ) )X (.\ rs 9