PHY 101: Lecture Ideal Spring and Simple Harmonic Motion 10.2 Simple Harmonic Motion and the Reference Circle 10.3 Energy and Simple Harmonic Motion 10.4 The Pendulum 10.5 Damped Harmonic Motion 10.6 Driven Harmonic Motion and Resonance 10.7 Elastic Deformation 10.8 Stress, Strain, and Hooke’s Law

PHY 101: Lecture 10 Simple Harmonic Motion and Elasticity 10.1 Ideal Spring and Simple Harmonic Motion

Ideal Spring - 1

Ideal Spring - 2 The bottom drawing in the figure illustrates the spring being compressed The hand applies pushing force to the spring, and it again undergoes a displacement from its unstrained length

Ideal Spring - 3

Ideal Spring - 4

Ideal Spring - 5

Simple Harmonic Motion - 1 In the figure, an object of mass m is attached to a spring on a frictionless In part A, the spring has been stretched to the right, so it exerts the leftward-pointing force F x When the object is released, this force pulls it to the left, restoring it towards the equilibrium position Consistent with Newton’s first law, the moving spring has inertia and moves beyond the equilibrium positon, compressing the spring as in part B

Simple Harmonic Motion - 2 The force exerted by the spring now points to the right The force brings the object to momentary halt Then it acts to restore the object to its equilibrium position Objects inertia again carries it beyond the equilibrium position stretching the spring and leading to the restoring force F x shown in Part C

Simple Harmonic Motion - 3 When the restoring force has the mathematical form given by F x = -kx, the type of friction-free motion is designated as simple harmomic motion”

Simple Harmonic Motion - 4 Attach a pen to the mass at end of spring Move a strip of paper past pen at a steady rate Shape of graph seems to be a sine or cosine function Maximum displacement is amplitude, A

PHY 101: Lecture 10 Simple Harmonic Motion and Elasticity 10.2 Simple Harmonic Motion and the Reference Circle

Simple Harmonic Motion and the Reference Circle - 1 Simple harmonic motion can be described in terms of displacement, velocity, and acceleration The figure is helpful in explaining these characteristics

Simple Harmonic Motion and the Reference Circle – 2 Small ball is attached to top of rotating turntable Ball moves in uniform circular motion Path is reference circle Shadow of moving ball falls on strip of film Film moves up at steady rate Shadow is recorded

Simple Harmonic Motion and the Reference Circle – 3 Figure is reference circle radius A Shows how to determine displacement on film At time t = 0, ball is at x = +A Ball move through angle  radians in a time t Ball moves with constant angular velocity  rad/s Angle is  =  t Displacement x of shadow is projection of radius A onto the x- axis x = Acos  = Acos  t

Simple Harmonic Motion and the Reference Circle – 4 Graph of x = Acos  t As time passes, shadow of ball oscillates from x=+A and x=-A Radius A of reference circle is the amplitude of the simple harmonic motion

Simple Harmonic Motion and the Reference Circle – 4 Ball moves one revolution around reference circle Shadow does one cycle of back-and-forth motion Time for 1 cycle is period T  =  /  t = 2  /T Frequency f is number of cycles of motion per second f = 1/T  = 2  /T = 2  f

SHM (Spring) Motion Properties Displacement  Vector distance of an object in SHM from its equilibrium position  Displacement can be either +x or –x, which indicates direction Amplitude (A)  Maximum displacement Period (T)  Time it takes the object to complete one cycle of motion  SI Units are second/cycle Frequency (f)  The number of cycles per second  SI Units are cycles/second or Hertz (Hz)  f = 1/T Angular Frequency (  )  The number of radians per second   = 2  f

SHM Properties – Example A 1.0-kg toy oscillating on a spring completes a cycle every 0.50 s What is the frequency of this oscillation?  T = 0.50 s  f = 1/T = 1/0.5 = 2.0 cycles/s or 2.0 Hz

Mass-Spring System Newton’s Second Law F = ma F is the spring force –kx m is mass attached to spring a is acceleration of spring in x-direction ma = -kx Solution of this calculus equation gives equations of motion of spring

Mass-Spring System Equations of Motion x = Acos(  t) + Bsin(  t) v = -A  sin(  t) +B  cos(  t) a = -A  2 cos(  t) - B  2 sin(  t) A and B depend on x and v at t = 0 -kx = F = ma -kAcos(  t) + Bsin(  t) = m[-A  2 cos(  t) - B  2 sin(  t)]

Mass-Spring System Equations of Motion When t = 0, x = +A When t = 0, x = -A When t = 0, x = 0, moving in + x-direction When t = 0, x = 0, moving in – x-direction

PHY 101: Lecture 10 Simple Harmonic Motion and Elasticity 10.3 Energy and Simple Harmonic Motion

Mass – Spring System Energy - 1  Potential Energy of spring  PE spring = ½ kx 2 Object of mass m attached to spring also has kinetic energy  E = KE + PE spring = ½ mv 2 + ½ kx 2 At maximum displacement (x = +A, or –A) the instantaneous velocity is zero (v = 0)  E = ½ m(0) 2 + 1/2 k(A) 2 = ½ kA 2  E = ½ kA 2 Total Energy of a mass in SHM on a spring

Mass – Spring System Energy - 2  E = KE + PE spring = ½ mv 2 + ½ kx 2 = 1/2kA 2  Solve for v

Mass-Spring System Velocity of Mass - 1 Velocity of an object in SHM Maximum speed of mass on a spring

Mass-Spring System Velocity of Mass - 2 Maximum speed of mass on a spring Amplitude, A, is equivalent to radius R, so

Velocity of Mass - Example An object of mass 0.50 kg is attached to a spring with spring constant 10 N/m The object is pulled down m from the equilibrium position and released What is its maximum speed?  v max = (A)sqrt(k/m)  v max = (0.050)sqrt(10/0.5) = 0.22 m/s

Mass Spring-System Example 1 A 0.50-kg mass oscillates on a spring with spring constant = 200 N/m What are (a) the period and (b) the frequency of the oscillation?  (a)  T = 2 x sqrt(m/k) = 2 x sqrt(0.50/200) = 0.31 s  (b)  f = 1/T = 1/0.31 = 3.2 cycles/s or 3.2 Hz

Mass-Spring System Example 2 Equation of motion for an oscillator in vertical SHM is given by y = (0.10 m) sin(100t) What are the (a) amplitude, (b) frequency, and (c) period of this motion? The equation of motion is  (a)  The amplitude is simply (0.10 m)  (b)  f = (1/2  ) x sqrt(k/m)  f = (1/2  ) x 100 = 15.9 Hz (cycles/s)  (c)  T = 1/f = 1/15.9 = s/cycle