Oscillations & Simple Harmonic Motion Oscillations Simple - - PDF document

Oscillations & Simple Harmonic Motion

• Oscillations
• Simple Harmonic Motion
• Mass-Spring System
• Angular Frequency
• Amplitude and Phase
• Displacement, Velocity & Acceleration in SHM
• Homework

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Oscillations

• Any motion that repeats itself in equal inter-

vals of time is called periodic or harmonic motion.

• The period, T, is the time required for one

complete oscillation or cycle.

• The frequency, f, is the number of oscilla-

tions per unit time.

• The unit of frequency is Hertz (Hz)

– 1 Hz = 1 cycle/s

• The frequency is the reciprocal of the period

f = 1 T

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Simple Harmonic Motion

• When a particle is under the effect of a linear

restoring force, the resulting motion is a spe- cial type of oscillatory motion called simple harmonic motion.

• For example, a block attached to a spring on

a frictionless surface moves in simple har- monic motion.

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Mass-Spring System F = ma −kx = md2x dt2 d2x dt2 + k mx = 0 x(t) = A cos (ωt + φ)

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Mass-Spring System Constants x(t) = A cos (ωt + φ) dx dt = −ωA sin (ωt + φ) d2x dt2 = −ω2A cos (ωt + φ) d2x dt2 + k mx = 0 −ω2A cos (ωt + φ) + k mA cos (ωt + φ) = 0 ω =

• k

m The constants A and φ are arbitrary and depend

• n how the motion starts.

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Angular Frequency

• Suppose we increase t by 2π

ω

x = A cos

 ω  t + 2π

ω

  + φ  

= A cos (ωt + 2π + φ) = A cos (ωt + φ)

• The function repeats itself after a time 2π

ω , so

the period of the motion is T = 2π ω = 2π

• m

k

• The frequency of the motion is

f = 1 T = ω 2π = 1 2π

• k

m

• The angular frequency of the motion is

ω = 2πf = 2π T =

• k

m

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Amplitude and Phase x(t) = A cos (ωt + φ)

• A is called the amplitude of the motion
• (ωt + φ) is called the phase of the motion
• φ is called the phase constant

φ x=Acos

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Displacement, Velocity & Acceleration in SHM x = A cos (ωt + φ) xmax = A v = dx dt = −ωA sin (ωt + φ) vmax = ωA =

• k

mA a = dv dt = −ω2A cos (ωt + φ) amax = ω2A = k mA

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Displacement, Velocity & Acceleration in SHM (cont’d)

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Example A block with a mass of 0.680 kg is fastened to a spring with a spring constant of 65.0 N/m. The block is pulled a distance x = 0.110 m from its equilibrium position at x = 0 on a frictionless surface and released from rest at t = 0. Find (a) the angular frequency, (b) the frequency, (c) the period, (d) the amplitude, (e) the maximum speed, (f) the maximum acceleration, (g) the phase constant, and (h) the displacement func- tion for the motion.

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Example Solution (a) ω =

• k

m =

• 65.0 N/m

0.680 kg = 9.78 rad/s

(b) f = ω

2π = 9.78 rad/s 2π rad

= 1.56 Hz (c) T = 1

f = 1 1.56 Hz = 0.641 s

(d) xmax = A = 0.110 m (e) vmax = ωA = (9.78 rad/s) (0.110 m) = 1.08 m/s (f) amax = ω2A = (9.78 rad/s)2 (0.110 m) = 10.5 m/s2 (g) φ = 0 (h) x(t) = A cos (ωt + φ) x(t)=(0.110 m) cos [(9.78 rad/s) t]

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Homework Set 22 - Due Fri. Nov. 5

• Read Sections 12.1-12.2
• Answer Questions 12.3 & 12.4
• Do Problems 12.3, 12.4, 12.6 & 12.14

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