Ch12 Simple harmonic motion and elasticity.docx

SIMPLE HARMONIC MOTION AND ELASTICITY Elasticity refers to the ability of an object to be deformed from its original shape by a force and return to its original shape when the deforming force is removed. Springs, for example, are coils of some material(steel or plastic for example) which do this very well if the deformation is not too large. When the deformation is performed within the elastic limits of the spring, the maximum deforming force is directly proportional to the displacement of the spring from equilibrium and obeys the equation: Fmax = kxmax where Fmax = the applied force, xmax = the maximum displacement from the equilibrium position, and k = the spring constant expressed in newtons per meter. The spring constant may be expressed in other units but it must be in N/m if it is used to calculate a force or energy in metric units. k is sometimes referred to as the stiffness of the spring since a larger k requires a larger force to produce the same displacement. When a spring is deformed, the spring exerts a force that opposes the deforming force. The force exerted by the spring is always directed towards its equilibrium position. Since the direction of the spring force is in the direction opposite to the deformation, the spring equation in vector form is: Fmax = - kxmax This is the mathematical form of Hooke's Law which says that the restoring force in a spring is directly proportional to the magnitude of the deformation and in the opposite direction to the deformation. Example Find the magnitude of the force required to stretch a spring with a spring constant of 248 N/m 3.00 cm. The diagram illustrates the relationship between spring deformation and force supplied by the spring. Example A 0.70 kg block is hung from a spring attached to the ceiling. A second block is now attached and the distance the spring is stretched triples. Find the mass of the second block. If the surface in the diagram above is frictionless, the mass on the end of the spring will repeat the same motion over and over. This type of movement is called simple harmonic motion and can be described mathematically. If the position of the mass at the end of the spring is plotted on the Y axis and time is plotted on the X axis, the resulting graph is called sinusoidal since it will have the shape of a trignometric sine or cosine function. The same type of motion can be observed by using a mass hanging vertically on the end of a spring. The equilibrium position occurs when mg = kdo where mg is the weight of the mass, k is the spring constant, and do is the distance the spring is deformed by the weight. Thus do is the equilibrium position for a spring and mass hanging vertically. The amplitude of oscillation then is the amount of deformation that occurs relative to do. The larger the deformation, the larger the amplitude. Since the motion of the mass can be described with a trignometric function, the time of motion must be related to an angular displacement. The equation is: x = Acos? = Acos?t where x is the displacement as a function of time, A is the amplitude of the motion, ? is the angular velocity as it relates to the trignometric unit circle, and t is the variable time. The time required to complete one cycle is called the period, T. It is related to the angular speed, ?, by the relationship ? = ??/?t. If ?? = 2?, then ?t = the period since that will be the time for one complete oscillation. We usually refer to the frequency rather than the period. The frequency is the number of cycles per second and is the reciprocal of the period which is the number of seconds per cycle. The unit of frequency is the hertz(Hz) which is one cycle per second. The angular speed is then 2?f and is sometimes called the angular frequency. Example A 0.80 kg object is attached at the end of a spring and set into simple harmonic motion. A graph of the motion is shown above.. Find (a) its amplitude, (b) its angular frequency, (c) the spring constant, (d) the speed of the object at t = 1 sec. and (e) the acceleration of the object at t = 1 sec. 17. REASONING AND SOLUTION a. Since the object oscillates between \* mergeformat, the amplitude of the motion is \* mergeformat. b. From the graph, the period is \* mergeformat. Therefore, according to Equation 10.4, \* mergeformat c. Equation 10.11 relates the angular frequency to the spring constant: \* mergeformat. Solving for k we find \* mergeformat d. At \* mergeformat, the graph shows that the spring has its maximum displacement. At this location, the object is momentarily at rest, so that its speed is . e. The acceleration of the object at \* mergeformat is a maximum, and its magnitude is \* mergeformat For an object oscillating along the X axis, its linear velocity is the x component of the tangential velocity of its projection onto the unit circle. Since the tangential velocity is related to the angular velocity by the relationship vt = r?, and r = A, the amplitude of motion on the unit circle, vt = A?. Since the linear velocity of the oscillating object is v = -vtsin?, we can write the equation for the linear velocity of the object as: v = -A?sin?t The negative sign must be added since the velocity is to the left when sin?t is positive and to the right when sin?t is negative. Maximum speed occurs when sin?t = 1(max value for sine function) so it is vmax = A?(magnitude only). Not all oscillatory motion is simple harmonic motion. There must be a restoring force directed towards the equilibrium position and the linear velocity must obey the equation shown above. Linear acceleration of an object in simple harmonic motion is not constant since the force causing the acceleration is not constant. Using calculus, it can be shown that the equation for acceleration for an object in simple harmonic motion is: a = -A?2cos?t The maximum value occurs when cos?t = 1. This happens when sin?t = 0 therefore when the velocity is 0. It is possible to find the frequency of vibration of an object at the end of a spring using these equations and Newton's second law. The net force is -kx so Newton's law becomes: -kx = ma By substituting x = Acos?t and a = -A?2cos?t, we get: -k(Acos?t) = m(-A?2cos?t) By dividing out -Acos?t and solving for ? we get: _____ ? = ?(k/m) _____ 2?f = ?(k/m) _____ f = (1/2?) ?(k/m) This is the equation used to determine the frequency of oscillation for a simple harmonic oscillator of known mass and spring constant. Example A spring stretches .018 m when a 2.8 kg object is hung from it. What total mass should be attached to make it oscillate at 3.0 Hz? 21. \* mergeformat REASONING The frequency of vibration of the spring is related to the added mass m by Equations 10.6 and 10.11: \* mergeformat (1) The spring constant can be determined from Equation 10.1. SOLUTION Since the spring stretches by 0.018 m when a 2.8-kg object is suspended from its end, the spring constant is, according to Equation 10.1, Solving Equation (1) for m, we find that the mass required to make the spring vibrate at 3.0 Hz is \* mergeformat ________________________________________________________________________ Objects which may be deformed like a spring can have elastic potential energy. The amount of work done to cause the deformation can be stored as potential energy and then later released as kinetic energy. The formula for calculating elastic potential energy of an ideal spring is: PE = ½kxf2 - ½kxo2 where k is the spring constant in N/m and x represents the initial and final distance that the spring is stretched. This also represents the work done on the spring by an outside force. If the spring is initially undeformed then the potential energy in the spring is simply ½kx2. Example An archer pulls back on the bow string 0.470 m before releasing the arrow. The spring constant for the bow is 425 N/m. Find (a) the potential energy stored in the drawn bow and (b) the speed of the 0.0300 kg arrow when it leaves the bow if all of this energy is converted into energy of motion of the arrow. 24. REASONING AND SOLUTION a. PE = (1/2) kx2 = (1/2) (425 N/m)(0.470 m)2 = \* mergeformat (10.13) b. Assuming that mechanical energy is conserved, KE = PE gives The total mechanical energy of an object consists of translational kinetic energy plus rotational kinetic energy plus gravitational potential energy plus elastic potential energy. The equation is: E = ½mv2 + ½I?2 + mgh + ½kx2 This total mechanical energy is constant if no nonconservative external forces(such as friction) do work. In this case, the final total energy is equal to the initial total energy. In general, some of the terms in the energy equation above will not be present in the initial energy equation and others will not be present in the final energy equation. If all of the quantities present except one are known or can be found, the equation can be solved for the one unknown quantity. Example A 2.00 kg object is hanging from the end of a spring with a spring constant of 50.0 N/m. It is pulled an additional 0.200 m down and released from rest. Complete the table below by finding the KE, Gravitational PE, elastic PE, and total energy for h = 0, 0.20m, and 0.40m. h KE \* mergeformat \* mergeformat E 0.000 m 0.200 m 0.400 m 25. \* mergeformat REASONING AND SOLUTION If we neglect air resistance, only the conservative forces of the spring and gravity act on the ball. Therefore, the principle of conservation of mechanical energy applies. When the 2.00 kg object is hung on the end of the vertical spring, it stretches the spring by an amount x, where \* mergeformat (10.1) This position represents the equilibrium position of the system with the 2.00kg object suspended from the spring. The object is then pulled down another 0.200 m and released from rest (\* mergeformat m/s). At this point the spring is stretched by an amount of \* mergeformat. This point represents the zero reference level (\* mergeformatm) for the gravitational potential energy. h = 0 m: The kinetic energy, the gravitational potential energy, and the elastic potential energy at the point of release are: \* mergeformat The total mechanical energy \* mergeformat at the point of release is the sum of the three energies above: \* mergeformat. h = 0.200 m: When the object has risen a distance of \* mergeformat above the release point, the spring is stretched by an amount of \* mergeformat. Since the total mechanical energy is conserved, its value at this point is still \* mergeformat. The gravitational and elastic potential energies are: \* mergeformat \* mergeformat Since \* mergeformat, \* mergeformat h = 0.400 m: When the object has risen a distance of \* mergeformat above the release point, the spring is stretched by an amount of \* mergeformat. At this point, the total mechanical energy is still \* mergeformat. The gravitational and elastic potential energies are: \* mergeformat \* mergeformat The kinetic energy is The results are summarized in the table below: h KE \* mergeformat \* mergeformat E 0.000 m 0.00 J 0.00 J 8.76 J 8.76 J 0.200 m 1.00 J 3.92 J 3.84 J 8.76 J 0.400 m 0.00 J 7.84 J 0.92 J 8.76 J A simple pendulum is a device that consists of a mass attached to a string of length L and insignificant mass. It is displaced from its equilibrium position and allowed to swing back and forth about that position. If the angle ? is small, the pendulum motion approximates that of a simple harmonic oscillator and the equations of an SHM can be used. The restoring force is caused by the component of weight perpendicular to the string, L, and points nearly at the equilibrium position. The resoring force also increases as the distance from the equilibrium position increases since a larger component of mg becomes perpendicular to the string. At the equilibrium position the restoring force is zero and the maximum velocity of the pendulum mass is reached. At the maximum displacement the velocity of the mass is zero and maximum acceleration occurs. Since a simple pendulum meets all of these conditions, we can approximate its behavior very closely with the SHM equations. Example Astronauts on a distant planet set up a simple pendulum 1.2 m long. They notice it takes 280 seconds to make 100 complete vibrations. Find the acceleration due to gravity on that planet. 41. \* mergeformat REASONING AND SOLUTION Recall that the relationship between frequency f and period T is \* mergeformat. Then, according to Equations 10.6 and 10.16, the period of the simple pendulum is given by \* mergeformat where L is the length of the pendulum. Solving for g and noting that the period is T = (280 s)/100 = 2.8 s, we obtain \* mergeformat Damped harmonic motion occurs when a frictional force does negative work on the system as it oscillates. This damping causes the amplitude of the motion to become smaller and smaller until it virtually disappears. The example shows the effect shock absorbers have on the up and down motion of a car when the front is lifted and released. Resonance is the result of driven harmonic motion. When a force acts on an oscillating system so that the force is in the same direction as the motion of the system, positive work is done on the system and the amplitude of the motion increases to reflect an increase in energy. An example of a system is a child(or adult) on a swing. In order for the person in the swing to go higher(greater amplitude) a force must be exerted on the swing as the swing begins to move forward. Successive pushes in this manner will result in a larger amplitude and a higher speed at the low point of the path. A push that opposes the motion of the swing acts as a damping force and reduces the amplitude of the motion.