# Chapter 15– Oscillations I.Simple harmonic motion (SHM) - Velocity - Acceleration II. Force law for SHM - Simple linear harmonic oscillator - Simple linear.

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Chapter 15– Oscillations I.Simple harmonic motion (SHM) - Velocity - Acceleration II. Force law for SHM - Simple linear harmonic oscillator - Simple linear harmonic oscillator (block attached to spring moving in 1D) (block attached to spring moving in 1D) III. Energy in SHM IV. Angular harmonic oscillator - Simple angular harmonic oscillator - Simple angular harmonic oscillator (torsion pendulum) (torsion pendulum) V. Pendulum VI. SHM and uniform circular motion VII. Damped simple harmonic motion VIII. Forced oscillations and resonance

I. Simple harmonic motion Examples: boats, guitar strings, diaphragms in telephones, oscillations of air molecules that transmit the sensation of sound, oscillations of atoms in a solid that convey the sensation of temperature. In the “real world” the oscillations are damped  motion dies out gradually  Mechanical energy transferred to thermal (friction)

I. Simple harmonic motion - Frequency: 1 Hertz = 1 oscillation per second = 1s -1 Units: Number of oscillations completed each second Number of oscillations completed each second. T = oscillation period  time for a complete oscillation (cycle).

- Periodic / harmonic motion: Any motion that repeats itself in a particular way. Displacement: x m, ω, φ =const. Magnitude of the maximum displacement. Simple Harmonic Motion (SHM)  Periodic motion  sinusoidal (sine or cosine) function of time. Amplitude (x m ): Phase constant (φ): Angle that depends on the particle’s displacement and velocity at t=0. Phase of the motion: (ωt + φ)

- Periodic / harmonic motion: Any motion that repeats itself in a particular way. Particle oscillating  maximum speed at x=0 minimum speed at x= ± x m

Angular frequency (ω): After one period  x(T+t)=x(t) The cosine function first repeat itself after 2π = 1 period rad/sUnits: Velocity of SHM: Velocity amplitude (v m ): v m = ω·x m t

Acceleration of SHM: Acceleration amplitude (a m ): a m = ω 2 ·x m In SHM, the acceleration is proportional to the displacement but opposite in sign and the two quantities are related by the square of the angular frequency.

Different amplitude Different period Different phase angle t t

II. Force law for SHM Restoring force similar to: Spring constant: mω 2 SHM is the motion executed by a particle of mass m subject to a force that is proportional to the displacement of the particle but opposite in sign. Linear simple harmonic oscillator Angular frequency: Period:

Kinetic energy: Mechanical energy: III. Energy in SHM Linear oscillator  ΔE mec = cte  back and forth energy transfer between kinetic and potential energy. Potential energy:

Mechanical energy:

IV. Angular Simple Harmonic oscillator Torsion pendulum If we rotate the disk some angular displacement θ from its rest position it will oscillate about that position in angular simple harmonic motion. Restoring torque: Torsion constant: κ (Greek letter kappa) (depends on length, diameter and material of suspension wire) Angular form of Hooke’s law Angular simple harmonic oscillator Period:

Angular simple harmonic motion Angular acceleration: α = - ω 2 θ Linear simple harmonic motion Linear acceleration: a = -ω 2 x Angular acceleration proportional to angular displacement from the equilibrium position but tends to rotate the system in the opposite direction to the displacement. Torque: Force: F = -k x Torque is proportional to the angular displacement but tends to rotate the system in the opposite direction.

V. Pendulum Simple pendulum Particle of mass “m” (pendulum’s bob) suspended from one end of an unstretchable massless string of length “L” that is fixed at the other end. Restoring torque about pendulum’s pivot point: Equilibrium position: θ = 0 “I” is the pendulum’s rotational inertia about the pivot point If θ is small  θ = sin θ SHM  α ~ - (cte)·θ In figure above, as the pendulum moves to right its acceleration to the left increases until it stops and starts moving to the left  The movement of a simple pendulum swinging through only small angles is approximately SHM.

Physical pendulum: Simple pendulum angular acceleration: α = - ω 2 θ = - (Lmg/I)· θ I =m L 2 rotational inertia of pendulum = simple pendulum  particle (bob) at radius L from pivot. Difference with respect to simple pendulum  the restoring force component F g sinθ has a moment arm of distance “h” about the pivot point rather than the string length L. Complicated distribution of mass about the pivot Small oscillation amplitude

“I” depends on shape of physical pendulum about O A physical pendulum will not swing if it pivots at its COM  h=0  T infinity  such pendulum will never complete one swing. Physical pendulum oscillating about O with period T can be represented by simple pendulum of length L 0 and same period. Center of oscillation: point along the physical pendulum at distance L 0 from O.

A physical pendulum can be used to measure “g”. Example : pendulum = rod of length L suspended from one end. h=L/2 (COM) I = I (COM)+M h 2 = ML 2 /12 + M (L/2) 2 = ML 2 /3 I (COM)=ML 2 /12

VI. Simple Harmonic Motion and Uniform Circular Motion Simple harmonic motion is the projection of uniform circular motion on a diameter of the circle in which the later motion occurs. Particle P’  uniform circular motion  φ’= ωt + φ Particle P  projection of particle P’ onto x-axis  x(t)= x m cos (ωt + φ) v = ω·R  v = ω·x m Projection of velocity of reference particle = velocity SHM Projection of radial acceleration of reference particle = acceleration SHM a r = ω 2 ·R  a = ω 2 ·x m

VII. Damped Simple Harmonic Motion Damped motion : when the motion of an oscillator is reduced by an external force. Example: Block oscillates vertically on a spring. From block, a rod extends to a vane submerged in liquid. Liquid exerts a damping force opposed to the motion  b = damping constant (kg/s) Assumption: Gravitational force on block is negligible compared to the damping force and the force on the block from the spring.

Solution: Frequency of damped oscillator: If b=0 (no damping) Mechanical energy  Displacement of damped oscillator Amplitude decreases exponentially with time Dampedoscillator: Un-damped oscillator:

VIII. Forced oscillations and resonance Examples: 1) Person swinging in a swing without anyone pushing  Free oscillation 2) Someone pushes the swing periodically  Forced or driven oscillations Driven oscillations Two frequencies: (1)Natural angular frequency ω of the system  when system oscillates freely after a sudden disturbance. a sudden disturbance. (2) External frequency ω d of the system  angular frequency of the external driving force causing the driven oscillations. If “rigid support” moves up and down  forced simple harmonic oscillator X m depends on ω d and ω Velocity of oscillations greatest when: ω d = ω  Resonance  Displacement amplitude greatest

An oscillator consists of a block of mass 0.500 kg connected to a spring. When set into oscillation with amplitude 35.0 cm, the oscillator repeats its motion every 0.500 s. Find the (a) period, (b) frequency, (c) angular frequency, (d) spring constant, (e) maximum speed, and (f) magnitude of the maximum force on the block from the spring.

Solution (a) The motion repeats every 0.500 s so the period must be T = 0.500 s. (b) The frequency is the reciprocal of the period: f = 1/T = 1/(0.500 s) = 2.00 Hz. (c) The angular frequency ω is ω = 2πf = 2π(2.00 Hz) = 12.6 rad/s. (d) The angular frequency is related to the spring constant k and the mass m by. We solve for k: k = m ω 2 = (0.500 kg)(12.6 rad/s)2 = 79.0 N/m. (e) Let x m be the amplitude. The maximum speed is v m = ω x m = (12.6 rad/s)(0.350 m) = 4.40 m/s. (f) The maximum force is exerted when the displacement is a maximum and its magnitude is given by F m = kx m = (79.0 N/m)(0.350 m) = 27.6 N.

In the figure below, two identical springs of spring constant 7580 N/m are attached to a block of mass 0.245 kg. The block is set oscillating on the frictionless floor. What is the frequency of oscillation?

Solution When displaced from equilibrium, the net force exerted by the springs is –2kx acting in a direction so as to return the block to its equilibrium position (x = 0). Since the acceleration a = d 2 x/dt 2, Newton’s second law yields Substituting x = x m cos( ω t + φ) and simplifying, we find

A block weighing 14.0 N, which can slide without friction on an incline at angle θ = 40.0°, is connected to the top of the incline by a massless spring of unstretched length 0.450 m and spring constant 120 N/m. (a) How far from the top of the incline is the block's equilibrium point? (b) If the block is pulled slightly down the incline and released, what is the period of the resulting oscillations?

Solution (a) the equilibrium position when the block is gently lowered until forces balance. If the amount the spring is stretched is x, then we examine force-components along the incline surface and find at equilibrium, The distance from the top of the incline is therefore (0.450 + 0.075) m = 0.525 m. (b) Just as with a vertical spring, the effect of gravity (or one of its components) is simply to shift the equilibrium position; it does not change the characteristics (such as the period) of simple harmonic motion.

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