Simple Harmonic Motion and Elasticity

Force is proportional to displacement For small displacements, the force required to stretch or compress a spring is directly proportional to the displacement x, or The constant k is called the spring constant or stiffness of the spring. A spring behaves according to the above equation is said to be an ideal spring.

10.1 The Ideal Spring and Simple Harmonic Motion spring constant Units: N/m

10.1 The Ideal Spring and Simple Harmonic Motion Example-1: A Tire Pressure Gauge The spring constant of the spring is 320 N/m and the bar indicator extends 2.0 cm. What force does the air in the tire apply to the spring?

10.1 The Ideal Spring and Simple Harmonic Motion

Tire Pressure? Tire pressure varies from 2 bar to 2.2 bar. Or, 28 psi to 32 psi 1 bar = 100 kPa = 100 000 Pa = 100 000 N/m2 Pressure = Force per unit Area P = F/A What is the area of the Tire Pressure Gauge?

10.1 The Ideal Spring and Simple Harmonic Motion Conceptual Example-2: Are Shorter Springs Stiffer? A 10-coil spring has a spring constant k. If the spring is cut in half, so there are two 5-coil springs, what is the spring constant of each of the smaller springs?

Shorter Springs are Stiffer The spring constant of each 5-coil spring is 2k. Spring constant  1/# of coils k for 10 coils 2k for 5 coils 10k for 1 coil

Restoring Force To stretch or compress a spring, a force must be applied to it. In accord with Newton’s third law, the spring exert an oppositely directed force of equal magnitude. This reaction force is applied by the spring to the agent that does the pulling or pushing. The reaction force is also called a “restoring force”. Fx = - kx The reaction force always points in a direction opposite to the displacement of the spring from its unstrained length.

Restoring Force The restoring force of a spring can also contribute to the net external force.

10.1 The Ideal Spring and Simple Harmonic Motion HOOKE’S LAW: RESTORING FORCE OF AN IDEAL SPRING The restoring force on an ideal spring is

Why it is called Restoring Force An object of mass m is attached to a spring on a frictionless table. In part A, the spring has been stretched to the right, so the spring exerts the left-ward pointing force Fx. When the object is released, this force pulls it to the left, restoring it toward its equilibrium position.

Restoring Force causes Simple Harmonic Motion When the object is released, the restoring force pulls it to its equilibrium position. In accord with Newton’s first law, the moving object has inertia and coasts beyond the equilibrium position, compressing the spring as in part B. The restoring force exerted by the spring now points to the right and after bringing the object to a momentary halt, acts to restore the object to its equilibrium position. Since no friction acts on the object the back-and-forth motion repeats itself. When the restoring force has the mathematical form given by Fx = -kx, the type of friction-free motion is designated as “simple harmonic motion”.

Simple Harmonic Motion By attaching a pen to the object and moving a strip of paper past it at a steady rate, we can record the position of the vibrating object as time passes. The shape of this graph is characteristic of simple harmonic motion and is called sinusoidal.

Simple Harmonic Motion Simple harmonic motion, like any motion, can be described in terms of displacement, velocity, and acceleration.

10.2 Simple Harmonic Motion and the Reference Circle DISPLACEMENT

10.2 Simple Harmonic Motion and the Reference Circle Radius = A The displacement of the shadow, x is just the projection of the radius A onto the x-axis: Where  = t , the angular speed in rad/s

Angular Speed,   = /t rad/s = t rad Angular displacement () for one cycle is 2 rad in T. So,  = 2/T  = 2 Because  is directly proportional to the frequency ,  is often called the angular frequency.

10.2 Simple Harmonic Motion and the Reference Circle amplitude A: the maximum displacement period T: the time required to complete one cycle frequency f: the number of cycles per second (measured in Hz)

10.2 Simple Harmonic Motion and the Reference Circle VELOCITY Tangential velocity, VT = Radius x angular velocity Velocity of the shadow, Vx = X component of VT Maximum Velocity of the shadow, Vx = A

10.2 Simple Harmonic Motion and the Reference Circle Example 3 The Maximum Speed of a Loudspeaker Diaphragm The frequency of motion is 1.0 KHz and the amplitude is 0.20 mm. What is the maximum speed of the diaphragm? Where in the motion does this maximum speed occur?

10.2 Simple Harmonic Motion and the Reference Circle The maximum speed occurs midway between the ends of its motion.

Simple harmonic motion? Is the motion of the lighted bulb simple harmonic motion, when each lights for 0.5s in sequence?

10.2 Simple Harmonic Motion and the Reference Circle ACCELERATION The ball on the reference circle moves in uniform circular motion, and, therefore, has centripetal acceleration ac that points toward the centre of the circle. The acceleration ax of the shadow is the x component of the centripetal acceleration ac .

Maximum Acceleration A loudspeaker vibrating at 1kHz with an amplitude of 0.20mm has a maximum acceleration of amax = A2 = 7.9 x 103 m/s2 Maximum acceleration occurs when the force acting on the diaphragm is a maximum. The maximum force arises when the diaphragm is at the ends of its path.

Frequency of Vibration With the aid of the Newton’s second law , it is possible to determine the frequency at which an object of mass m vibrates on a spring. Mass of the spring is negligible Only force acting is the restoring force.

10.2 Simple Harmonic Motion and the Reference Circle FREQUENCY OF VIBRATION Larger spring constants k and smaller masses m result in larger frequencies.

10.2 Simple Harmonic Motion and the Reference Circle Example 6 A Body Mass Measurement Device The device consists of a spring-mounted chair in which the astronaut sits. The spring has a spring constant of 606 N/m and the mass of the chair is 12.0 kg. The measured period is 2.41 s. Find the mass of the astronaut.

10.2 Simple Harmonic Motion and the Reference Circle

10.3 Energy and Simple Harmonic Motion A compressed spring can do work.

Average magnitude of Force Spring force at x0 is kx0 Spring force at xf is kxf Average Fx = ½(kx0+kxf)

10.3 Energy and Simple Harmonic Motion Work done by the average spring force: Final elastic PE Initial elastic PE

10.3 Energy and Simple Harmonic Motion DEFINITION OF ELASTIC POTENTIAL ENERGY The elastic potential energy is the energy that a spring has by virtue of being stretched or compressed. For an ideal spring, the elastic potential energy is SI Unit of Elastic Potential Energy: joule (J)

10.3 Energy and Simple Harmonic Motion Conceptual Example 8 Changing the Mass of a Simple Harmonic Oscilator The box rests on a horizontal, frictionless surface. The spring is stretched to x=A and released. When the box is passing through x=0, a second box of the same mass is attached to it. Discuss what happens to the (a) maximum speed (b) amplitude (c) angular frequency.

Doubling the Mass of a Simple Harmonic Oscillator At x=0m, a second box of the same mass and speed vmax is attached. So, the max KE is doubled as mass is doubled. So, when the spring compresses it will have double the PEe . As PEe is doubled max amplitude will be 2 times

Doubling the Mass causes max amplitude to 2 times Before adding the second mass, the displacement x1 is related to the PE as: As PEe is doubled the new amplitude is x2

Doubling the Mass of a Simple Harmonic Oscillator Angular frequency is So, a doubling of mass will cause the angular frequency reduce to (1/ 2 ) 

10.3 Energy and Simple Harmonic Motion Example 9 A falling ball on a vertical Spring A 0.20-kg ball is attached to a vertical spring. The spring constant is 28 N/m. When released from rest, how far does the ball fall before being brought to a momentary stop by the spring?

10.3 Energy and Simple Harmonic Motion

A simple pendulum consists of 10.4 The Pendulum A simple pendulum consists of a particle mass m attached to a frictionless pivot by a cable of negligible mass.

simple pendulum The force of gravity is responsible for the back-and-forth rotation about the axis at P. A net torque is required to change the angular speed. The gravitational force mg produces the torque. = - (mg)l - sign to represent the restoring force = - (mg)L = - k’  where k’ = mgL Same as F = - kx So,  = (k’/m) = (mgL/m) = (mgL/I) in rotational motion in place of mass moment of inertia ‘I’ will appear

simple pendulum So,  = (k’/m) = (mgL/m) = (mgL/I) in rotational motion in place of mass moment of inertia ‘I’ will appear The moment of inertia of a particle of mass m, rotating at a radius L about an axis is I = mL2 Frequency of a simple pendulum is Which does not depend on the mass of the particle.

Determine the length of a simple pendulum that will 10.4 The Pendulum Example 10 Keeping Time Determine the length of a simple pendulum that will swing back and forth in simple harmonic motion with a period of 1.00 s.

10.5 Damped Harmonic Motion In simple harmonic motion, an object oscillated with a constant amplitude. In reality, friction or some other energy dissipating mechanism is always present and the amplitude decreases as time passes. This is referred to as damped harmonic motion.

10.5 Damped Harmonic Motion simple harmonic motion 2 & 3) underdamped critically damped 5) overdamped

10.6 Driven Harmonic Motion and Resonance When a force is applied to an oscillating system at all times, the result is driven harmonic motion. Here, the driving force has the same frequency as the spring system and always points in the direction of the object’s velocity.

10.6 Driven Harmonic Motion and Resonance Resonance is the condition in which a time-dependent force can transmit large amounts of energy to an oscillating object, leading to a large amplitude motion. Resonance occurs when the frequency of the force matches a natural frequency at which the object will oscillate.

Elastic Deformation All materials become distorted when they are squeezed or stretched. Those materials return to their original shape when the deforming force is removed, such materials are called “elastic”.

Atomic View of Elastic materials 10.7 Elastic Deformation Atomic View of Elastic materials Because of these atomic-level “springs”, a material tends to return to its initial shape once forces have been removed. ATOMS FORCES

Stretching force Magnitude of the deforming force can be expressed as follows, provided the amount of stretch or compression is small compared to the original length of the object.

STRETCHING, COMPRESSION, AND YOUNG’S MODULUS 10.7 Elastic Deformation STRETCHING, COMPRESSION, AND YOUNG’S MODULUS Y is the proportionality constant called Young’s modulus Young’s modulus has the units of pressure: N/m2

Young’s modulus The magnitude of the force is proportional to the fractional increase (or decrease) in length L/L0, rather than the absolute change L. Young’s modulus depends on the nature of the material. For a given force, the material with higher Y undergoes the smaller change in length. This force also called the “tensile” force, because they cause a tension in the material

10.7 Elastic Deformation

Example 12 Bone Compression 10.7 Elastic Deformation Example 12 Bone Compression In a circus act, a performer supports the combined weight (1080 N) of a number of colleagues. Each thighbone of this performer has a length of 0.55 m and an effective cross sectional area of 7.7×10-4 m2. Determine the amount that each thighbone compresses under the extra weight.

10.7 Elastic Deformation

Shear Deformation It is possible to deform a solid object in a way other than stretching or compressing it. When a top cover of a book is pushed the pages below it become shifted relative to the stationary bottom cover. The resulting deformation is called a shear deformation.

Shear Deformation When a force F is applied parallel to the surface of an object of area A, the object experiences a shear x (the top surface moves relative to the bottom surface) for an object with thickness L0. The magnitude of the shearing force is proportional to the fractional shear x in thickness x/L0, rather than the absolute change x.

Shear Modulus The constant of proportionality S is called the shear modulus. Shear modulus depends on the nature of the material For a given force, the material with higher S undergoes the smaller shear x . Shear modulus has the units of pressure: N/m2

Shear Deformation And The Shear Modulus

SHEAR DEFORMATION AND THE SHEAR MODULUS 10.7 Elastic Deformation SHEAR DEFORMATION AND THE SHEAR MODULUS The shear modulus has the units of pressure: N/m2

10.7 Elastic Deformation

Example 14: Shear Modulus of J-E-L-L-O 10.7 Elastic Deformation Example 14: Shear Modulus of J-E-L-L-O You push tangentially across the top surface with a force of 0.45 N. The top surface moves a distance of 6.0 mm relative to the bottom surface. What is the shear modulus of Jell-O?

Shear Modulus of J-E-L-L-O 10.7 Elastic Deformation Shear Modulus of J-E-L-L-O Jell-o can be deformed easily, because its Shear Modulus is significantly ___________ than that of other rigid materials, e.g., steel

Young’s modulus vs. Shear modulus Young’s modulus (Y) refers to the change in length of one dimension of a solid object as a result of tensile or compressive force. The shear modulus (S) refers to a change in shape of a solid object as a result of shearing force.

Young’s modulus vs. Shear modulus The tensile force is perpendicular to the surface of area A The shear force is parallel to the surface The distances L and L0 (length) are parallel The distances  x and L0 (thickness) are perpendicular

Volume Deformation When applied compressive forces changes the size of every dimension (length, width, and depth), leading to a decrease in volume, the resulting deformation is called a volume deformation. This kind of overall compression occurs when an object is submerged in a liquid. The force acting in such situation s are applied perpendicular to every surface. The perpendicular force per unit area is pressure.

Pressure The pressure P is the magnitude of the force F acting perpendicular to a surface divided by the area A over which the surface acts: SI unit of Pressure is Pa (pascal) = N/m2 F A

Volume Deformation The change in pressure P needed to change the volume by an amount V is directly proportional to the fractional change V/V0 in volume: Change in pressure P = final pressure P - initial pressure P0 Change in volume v = final volume V - initial volume V0

VOLUME DEFORMATION AND THE BULK MODULUS 10.7 Elastic Deformation VOLUME DEFORMATION AND THE BULK MODULUS B is the proportionality constant called Bulk modulus The Bulk modulus has the units of pressure: N/m2 The negative sign represents the increase in pressure decreases the volume

10.7 Elastic Deformation

Stress, Strain, and Hooke’s Law All these modified equations specify the amount of force needed per unit area for a given amount of elastic deformation. In general, the ratio of the magnitude of the force to the area is called the stress. The right side of each equation involves the change in a quantity (L, X, or V) divided by a quantity (L or V) relative to which the change is compared. Each of these ratios are referred to as the strain that results from the stress.

Stress, Strain, and Hooke’s Law Stress and strain are directly proportional to one another. This relationship was first discovered by Robert Hooke and referred to a Hooke’s Law.

10.8 Stress, Strain, and Hooke’s Law In general the quantity F/A is called the stress. The change in the quantity divided by that quantity is called the strain: HOOKE’S LAW FOR STRESS AND STRAIN Stress is directly proportional to strain. Strain is a unitless quantitiy. SI Unit of Stress: N/m2

10.8 Stress, Strain, and Hooke’s Law In reality, materials obey Hooke’s law only up to a certain limit, proportionality limit. Elastic limit is the point beyond which the object no longer returns to its original size and shape when the stress is removed.

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