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Vibrations and waves

Hooke’s Law

The force exerted on an object attached to a spring is proportional to the displacement of the spring.

kxF −=The constant of proportionality is the spring’s constant.

The force acts in opposite direction to the displacement.

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Acceleration of the Object on a spring

Hooke’s Law gives the force

kxF −=

Newton’s law related the force with the acceleration

maF =

Thus

xm

k

m

Fa −==

Hooke’s Law: example

A 0.35 Kg object attached to a spring of constant 130 N/m is free to move on a frictionless surface. If the object is released from rest at x=0.10 m, find the force on it and it’s acceleration for

a) x=0.10 mb) x=0.05 mc) x=0 md) x=-0.05 me) Sketch a graph of F(x)

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Simple Harmonic Motion

Simple harmonic motion is the type of motion that results from a force that is proportional to the displacement but opposite direction. It is periodic.

Amplitude A: magnitude of the maximum position of the object relative to its equilibrium position. In the simple harmonic motion the system will oscillate between +A and –A.

Period T: the time it takes for the object to complete a full cycle. If the system starts at the maximum x=+A, a full cycle involves going all the way to x=-A and back to x=+A again.

Frequency f: the number of cycles completed per unit time.

Simple Harmonic motion: question time

A block on the end of a spring is pulled to position x=A and released. Through what total distance does it travel in a full cycle?

a) A/2 b) A c) 2A d) 4A

In a simple harmonic motion the acceleration and the displacement point in the same direction.

(a)True or (b) false

Two kids throw a ball to and fro repeating the same path over and over again.

Is the motion of the ball periodic? (a) yes, (b) no. Is it a simple harmonic motion? (a) yes, (b) no.

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Energy of the object and spring system

The potential energy stored in a spring is proportional to the square of the displacement

a) Potential energy =0b) Kinetic energy is

being converted into potential energy

c) Potential energy is maximum

2

2

1kxPE =

Velocity in simple harmonic motion

One can determine the velocity of the object from energy conservation

222

2

1

2

1

2

1kxmvkA +=

)( 22 xAm

kv −±=

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Energy and velocity in SHM: example

A 0.35 Kg object attached to a spring of constant 130 N/m is free to move on a frictionless surface. The object is released from rest at x=0.10 m.

a) Calculate the total energy of the systemb) Calculate the maximum speed of the objectc) What is the velocity when the displacement is 0.05?d) Compute the kinetic energy and the potential energy for

x=0.10 m, x=0.05 m, x=0 m, x=-0.05, x=-0.01. e) Sketch a graph of KE(x) and PE(x)

Comparing SHM with UCM: derivation

The projection of the uniform circular motion (UCM) in one of the axes reproduces the pattern of the simple harmonic motion (SHM). Knowing the definition of the period in UCM, prove that the period in SHM is given by

0

2

v

AT

π=

20

2

2

1

2

1mvkA =

k

mT π2=

m

kf == πω 2

One can easily derive the expression for the angular frequency

Useful relations

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Period and frequency: question time

Consider a mass M attached to a spring, with a spring constant K, oscillating with period T on a frictionless surface with amplitude A. You now replace that mass by another with 2M and set it into oscillation with amplitude A. What happens to the period?

a) 4T b) 2T c) T/2 d) none are correct

Did the total energy of the system change?(a) yes, (b) no

Period and frequency: example

A 1300 Kg car is constructed on a frame supported by four springs, each with k=20000N/m. If two people riding in the car have a combined mass of 160 Kg, find the frequency of free vibration of the car when it is driven over a pothole in the road. How long does it take to complete 3 periods?

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Position, velocity and acceleration

θcosAx =

tωθ =

)cos( tAx ω=

Period and frequency: example

Consider the system represented in the figure and determine which of the following statements is correct.

a) If the amplitude of the oscillation of the mass is doubled, the total energy is doubled.

b) If the amplitude is doubled, the maximum acceleration is doubled.

c) In order to half the angular frequency of oscillation, one has to double the mass.

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Motion of a pendulum: derivation

Consider the forces in a pendulum. Taking into account the small angle approximation derive the expression for the period of a pendulum:

θθ =sin

g

LT π2=

small angle approximation

The pendulum versus mass and spring system

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Pendulum: example

x

We want to measure the height of a very tall tower using a cable attached to the roof. We find that it takes 15 s for a 30 kg mass attached to the end of the cable to performed a complete oscillation. Estimate the height of the tower.

Waves

What is a wave?An oscillation that is both a function of time and of distance.

Snapshot: behaviour with x

Looking at one point only behaviour with t

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Waves

A function of timeand of distance.

Types Waves

Transverse: oscillation is perpendicular to the direction of propagation.

Longitudinal: oscillation is in the direction of the propagation

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Wavelength

The wavelength is the distance between two maximum points of the wave.

Tt

xv

λ=∆∆=The wave speed is

λ

λ

x

Wavelength: example

x

A wave traveling in the position x has frequency 25.0 Hz. Find the amplitude, the wavelength, the period, and the speed of the wave.

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Interference of waves

x

Interference of waves

Superposition principle:if two or more traveling waves are moving through a medium, the resulting wave is found by adding together the displacements of the individual waves point by point

x

What happens when wave a) interferes with b) in both cases?

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Interference of waves

x

Constructive interference Destructive interference

Reflection of waves

Rigid boundary Free boundary