Oscillations and periodic motion

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# Oscillations and periodic motion - PowerPoint PPT Presentation

Oscillations and periodic motion. Periodic Motion Motion that repeats itself over a fixed and reproducible period of time is called periodic motion .

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### Oscillations and periodic motion

Periodic Motion
• Motion that repeats itself over a fixed and reproducible period of time is called periodic motion.
• The revolution of a planet about its sun is an example of periodic motion. The highly reproducible period (T) of a planet is also called its year.
• Mechanical devices on earth can be designed to have periodic motion. These devices are useful timers. They are called oscillators.
Simple Harmonic Motion
• You attach a weight to a spring, stretch the spring past its equilibrium point and release it. The weight bobs up and down with a reproducible period, T.
• Plot position vs time to get a graph that resembles a sine or cosine function. The graph is “sinusoidal”, so the motion is referred to as simple harmonic motion.
• Springs and pendulums undergo simple harmonic motion and are referred to as simple harmonic oscillators.
Equilibrium is where kinetic energy is maximum and potential energy is zero.

Analysis of graph

Analysis of graph

Maximum and minimum positions have maximum potential energy and zero kinetic energy.

Oscillator Definitions
• Amplitude
• Maximum displacement from equilibrium.
• Related to energy.
• Period
• Length of time required for one oscillation.
• Frequency
• How fast the oscillator is oscillating.
• f = 1/T
• Unit: Hz or s-1
Springs
• Springs are a common type of simple harmonic oscillator.
• Our springs are “ideal springs”, which means
• They are massless.
• They are both compressible and extensible.
• They will follow Hooke’s Law.
• F = -kx
Fs = -kx

Review of Hooke’s Law

The force constant of a spring can be determined

by attaching a weight and seeing how far it stretches.

Fs

m

mg

m = 30 0 g = 0.3 kg

k = 25 N / m

• Calculate the period of a 300-g mass attached to an ideal spring with a force constant of 25 N/m.

T = 2π√(m/k)

= 0.69 s

T = 2π√(0.3 / 25)

A 300-g mass attached to a spring undergoes simple harmonic motion with a frequency of 25 Hz. What is the force constant of the spring?

m = 30 0 g = 0.3 kg

k = ? N / m

f = 25 Hz

T = 1 / f

T = 2π√(m/k)

1 / f = 2π√(m/k)

1 / f2 = 4π2m/k

k / f2 = 4π2m

k = 4π2mf2

k = 4π2 ∙ 0.3 ∙ (25)2

= 7402.2 N / m

An 80-g mass attached to a spring hung vertically causes it to stretch 30 cm from its unstretched position. If the mass is set into oscillation on the end of the spring, what will be the period?

m = 30 0 g = 0.3 kg

k = 25 N / m

T = 2π√(m/k)

= 0.69s

T = 2π√(0.3 / 25)

2 Springs act stronger than 1 Spring

If I move the block 10 cm, each spring must move 10 cm

Spring combinations

• Parallel combination: springs work together.
• Series combination: springs work independently

Double the work

If I move the block 10 cm, each spring must move 5 cm

2 Springs act weaker than 1 Spring

half the work

Question?
• Does this combination act as parallel or series?

10 cm

Stretch 10 cm

Compress 10 cm

Acts like parallel system

Conservation of Energy
• Springs and pendulums obey conservation of energy.
• The equilibrium position has high kinetic energy and low potential energy.
• The positions of maximum displacement have high potential energy and low kinetic energy.
• Total energy of the oscillating system is constant.
Sample problem.
• A spring of force constant k = 200 N/m is attached to a 700-g mass oscillating between x = 1.2 and x = 2.4 meters. Where is the mass moving fastest, and how fast is it moving at that location?

In the center at x = 1.8 m

Pendulums
• The pendulum can be thought of as a simple harmonic oscillator.
• The displacement needs to be small for it to work properly.
Period of a pendulum
• T: period (s)
• l: length of string (m)
• g: gravitational acceleration (m/s2)
Sample problem
• Predict the period of a pendulum consisting of a 500 gram mass attached to a 2.5-m long string.

T = ? s

a = 9.8 m/s2

l = 2.5 m

T = 2π√(l/g)

= 2π√(2.5/9.8)

= 3.17s

Sample problem
• Suppose you notice that a 5-kg weight tied to a string swings back and forth 5 times in 20 seconds. How long is the string?

f = 5/20 s = 0.25 Hz

a = 9.8 m/s2

l = ? m

T= 1/f = 1/0.25 = 4 s

T = 2π√(l/g)

T2 = 4π2l/g

= 4 m

= 0.42∙9.8 / (4π2)

l = T2 g / (4π2)