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Oscillations. Spring time again…. Let’s revisit Hooke’s law X>0 F x <0 Spring pulls back the mass toward its equilibrium position X=0 F x =0 Equilibrium position, no force X<0 F x >0 Spring pushes the mass back toward its equilibrium position. Restoring force and harmonic motion.

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Oscillations


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Presentation Transcript
spring time again
Spring time again…
  • Let’s revisit Hooke’s law
  • X>0 Fx<0
  • Spring pulls back the mass toward its equilibrium position
  • X=0 Fx=0
  • Equilibrium position, no force
  • X<0 Fx>0
  • Spring pushes the mass back toward its equilibrium position

PHY231

restoring force and harmonic motion
Restoring force and harmonic motion
  • A restoring force always pushes or pulls the object toward its equilibrium position
  • Hooke’s law is an example of restoring force
  • If the net force occurring on an object is along the direction of motion and is similar to Hooke’s law then the object follows a simple harmonic motion

PHY231

period of the motion
Period of the motion
  • The period T is the time it takes for the mass to come back to the exact same location with the exact same velocity. For the horizontal spring system
  • Mass comes back to same x and v periodically
  • Max displacement
  • t1=0 s t2=6.3 s
  • T= t2-t1 =6.3 s
  • Intermediate pos.
  • t1=2.5 s t2=8.8 s
  • T= t2-t1 =6.3 s

PHY231

frequency angular frequency
Frequency, angular frequency
  • The frequency f is the inverse of the period T. It corresponds to the number of cycles completed each second. For a mass on a spring

SI unit Hz=1/s

  • The angular frequency w is defined as 2p times the frequency

SI unit rad/s

PHY231

visualization

k =spring-constant , m=mass , w=ang. freq

w = (k/m)0.5

Visualization
  • Equilibrium position
    • Zero F
    • Zero accel
    • Zero SPE
    • Maximum KE = 0.5*kA2
    • Maximum velocity = Aw
  • Maximum displacement A
    • Maximum F = kA
    • Maximum accel = Aw2
    • Maximum SPE = 0.5*kA2
    • Zero KE
    • Zero velocity

A

A

time evolution
Time evolution

T

  • Simple harmonic motion is described by the following equations
  • Maximum values

T

Aw

Aw2

vertical mass spring system
Vertical mass-spring system
  • Exactly the same as horizontal mass-spring system but for the fact that gravitational potential energy is now also varying when the mass moves up and down
  • If x refers to the displacement with respect to the equilibrium position then all the previous formulas are still valid
at equilibrium position
At equilibrium position
  • When an object is moving in simple harmonic motion, which of the following is at a minimum when the displacement from equilibrium is zero?
  • A) the magnitude of the velocity
  • B) the magnitude of the acceleration
  • C) the kinetic energy
  • D) the total mechanical energy

PHY231

at equilibrium position1
At equilibrium position
  • A) the magnitude of the velocity
  • B) the magnitude of the acceleration
  • C) the kinetic energy
  • D) the total mechanical energy

PHY231

period t
Period T
  • Consider a mass-spring system on a horizontal frictionless surface set in simple harmonic motion with amplitude A. How does the period of the system change if the mass is quadrupled?
  • A) It doubles
  • B) It quadruples
  • C) It is half
  • D) It doesn’t change

PHY231

period t1
Period T

Consider a mass-spring system on a horizontal frictionless surface set in simple harmonic motion with amplitude A. How does the period of the system change if the mass is quadrupled?

  • A) It doubles
  • B) It quadruples
  • C) It is half
  • D) It doesn’t change

PHY231

cosine function
Cosine function
  • A mass on a spring vibrates in simple harmonic motion at a frequency of 4.0 Hz and an amplitude of 8.0 cm. If a timer is started when its displacement is a maximum (hence x = 8 cm when t = 0), what is the displacement of the mass when t = 3.7 s?
  • A) zero
  • B) 0.025 m
  • C) 0.036 m
  • D) 0.080 m

PHY231

cosine function1
Cosine function
  • A mass on a spring vibrates in simple harmonic motion at a frequency of 4.0 Hz and an amplitude of 8.0 cm. If a timer is started when its displacement is a maximum (hence x = 8 cm when t = 0), what is the displacement of the mass when t = 3.7 s?
  • A) zero
  • B) 0.025 m
  • C) 0.036 m
  • D) 0.080 m

PHY231

spring combo
Spring combo
  • A mass of 4.0 kg, resting on a horizontal frictionless surface, is attached on the right to a horizontal spring with spring constant 20 N/m and on the left to a horizontal spring with spring constant 50 N/m.
  • If this system is moved from equilibrium, what is the effective spring constant?
  • A) 30 N/m
  • B) -30 N/m
  • C) 70 N/m
  • D) 14 N/m

PHY231

spring combo1
Spring combo
  • A) 30 N/m
  • B) -30 N/m
  • C) 70 N/m
  • D) 14 N/m

Fright-spring

Fleft-spring

Equilibrium position

PHY231

pendulum
Pendulum
  • The pendulum is a mass swinging at the end of a light string
  • If the amplitude of the motion is small, then the motion of the pendulum is approximately a simple harmonic motion
  • For the mass-spring system

acceleration = - constant * displacement

We want to find an equivalent equation for the pendulum

pendulum period
Pendulum period
  • We have found for the pendulum
  • We see that it is similar in form to the spring mass equation a = - k/m*x
  • acceleration = - constant * displacement
  • We can use the results of the spring mass system if we replace:
  • x by q, a by a, and k/m by g/L
  • Thus, the period for the pendulum is given by

PHY231

motion pendulum spring mass
Motion: pendulum / spring-mass
  • We saw that both systems have similar equations
  • We can visualize the similarities in their motions

PHY231

demo pendulum physlet
Demo: Pendulum physlet
  • Dependence of T with the length of the pendulum
  • We try two different lengths for the pendulum
    • L1 = 0.50 m and L2=2.0 m
    • On earth g = 9.8 m/s2
    • T1=2*π*√(0.5/9.8 )= 1.4 s
    • T2=2*π*√(2.0/9.8 )= 2.8 s

PHY231

result with physlet
Result with physlet

L = 0.50 m

T = 1.4 s

L = 2.0 m

T = 2.8 s

different planet
Different planet
  • What should be the period of a pendulum on the moon if its period on earth is 2.0 s ?
  • gearth = 5.8*gmoon
  • A) 0.34 s
  • B) 0.83 s
  • C) 4.8 s
  • D) 12 s

PHY231

different planet1
Different planet
  • What should be the period of a pendulum on the moon if its period on earth is 2.0 s ?
  • gearth = 5.8*gmoon
  • A) 0.34 s
  • B) 0.83 s
  • C) 4.8 s
  • D) 12 s

PHY231

result with physlet1
Result with physlet

Earth

T = 2.0 s

Moon

T = 4.9 s