Oscillations

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# Oscillations - PowerPoint PPT Presentation

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

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## PowerPoint Slideshow about 'Oscillations' - catori

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### Oscillations

PHY231

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
• 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
• 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
• 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

PHY231

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

T

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

T

Aw

Aw2

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
• 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 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
• 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
• C) It is half
• D) It doesn’t change

PHY231

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
• C) It is half
• D) It doesn’t change

PHY231

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 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
• 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 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
• 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
• 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
• We saw that both systems have similar equations
• We can visualize the similarities in their motions

PHY231

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

L = 0.50 m

T = 1.4 s

L = 2.0 m

T = 2.8 s

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 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 physlet

Earth

T = 2.0 s

Moon

T = 4.9 s