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  • CHAPTER 13

    Simple Harmonic Motion

    1

  • 2

  • 3

  • Oscillations

    An object oscillates when it moves back and forth repeatedly, on either

    side of some equilibrium position. If the object is stopped from

    oscillating, it returns to the equilibrium position.

    Oscillations can be observed in different ways:

    1) Time keeping

    2) Planetary Motion

    3) Musical Instruments

    4

    Source: flickr

    Source: Pixabay

    Source: Pixabay

  • Natural (free) Oscillations

    The easiest oscillations to understand are natural

    oscillations, which every oscillator possesses. Every

    oscillator has a natural frequency π’‡πŸŽ of vibration, the frequency with which it vibrates freely after an initial

    disturbance.

    Simplest examples of natural oscillations: tuning fork and

    pendulum.

    5 https://commons.wikimedia.org/wiki/File:Newtons_cradle.gif

  • Forced Oscillations

    If a force is applied continuously to an oscillator, this will

    cause the oscillator to vibrate in a forced oscillation. The

    frequency of a forced oscillation is not the natural frequency

    of the oscillator but rather the forcing frequency of the

    external force.

    6

  • Simple Harmonic Motion

    A simple harmonic motion is a special kind of oscillations.

    Definition:

    A simple harmonic motion is defined as the motion of a

    particle about a fixed point such that its acceleration π‘Ž is proportional to its displacement π‘₯ from the fixed point, and is directed towards the point.

    7

  • Simple Harmonic Oscillators

    1) Mass – Spring System

    2) A Pendulum

    8Wikipedia Commons

    Source: Wikipedia Commons

  • Amplitude, Period and Frequency

    The displacement changes between positive

    and negative values.

    The maximum displacement is called Amplitude, π’™πŸŽ or A.

    The period, T is the time for one complete oscillation.

    The frequency,f is the number of oscillations per unit time,

    a reciprocal of 𝑇.

    𝑂: Equilibrium Position

    𝑋, π‘Œ:Extreme Points

    9

    X

    Y

  • Amplitude, Period and Frequency

    Recall:

    Frequency, 𝑓 = 1

    𝑇

    Angular frequency,

    πœ” = 2πœ‹π‘“ Or

    πœ” = 2πœ‹

    𝑇

    10

  • Example 1

    For the graph below, determine the amplitude, frequency and time

    period of the oscillation.

    Answer:

    Amplitude, 𝐴 = 0.8 m

    Time period, 𝑇 = 3 Γ— 2 = 6 s

    Frequency, 𝑓 = 1

    6

    = 1.67 Hz

    11

  • Simple Harmonic Motion

    A system will oscillate if

    there is a force acting on it

    that tends to pull it back to

    its equilibrium position – a

    restoring force.

    In a swinging pendulum the

    combination of gravity and

    the tension in the string that

    always act to bring the

    pendulum back to the centre

    of its swing.

    Tension

    Gravity

    Resultant

    (restoring

    force)

  • For a spring and mass the

    combination of the gravity

    acting on the mass and

    the tension in the spring

    means that the system will

    always try to return to its

    equilibrium position. M

    T

    g

    Restoring

    force

    Simple Harmonic Motion

  • kxF ο€­ο€½

    When the force acting on the particle is directly proportional

    to the displacement and opposite in direction, the motion is

    said to be simple harmonic motion.

    xF 

    Simple Harmonic Motion

  • Relation of SHM with Circular Motion

    The pattern of SHM is the same as the side view of an

    object moving at a constant speed around a circular path

    The amplitude of the

    oscillation is equal to

    the radius of the circle.

    The object moves quickest

    as it passes through the

    central equilibrium position.

    The time period of the

    oscillation is equal to the

    time taken for the object to

    complete the circular path.Source: Wikimedia Commons

  • General SHM Equations

    Most of the time, simple harmonic motion can be described by these two

    basic trigonometric functions:

    Where (πœ”π‘‘ + πœ™) denotes the phase of SHM. πœ” denotes the angular frequency and πœ™ is the phase constant.

    π‘₯ = π‘₯0 sin(πœ”π‘‘ + πœ™) or π‘₯ = π‘₯0 cos(πœ”π‘‘ + πœ™)

    16

  • Phase Difference

    The term phase describes the point that an oscillating mass has

    reached within the complete cycle of an oscillation.

    It is often important to describe the phase difference between two

    oscillations.

    17

  • Calculating Phase Difference

    To calculate the phase difference, follow these steps:

    1) Measure the time interval 𝑑 between two corresponding points on the graphs.

    2) Determine the period 𝑇 for one complete oscillation.

    3) Calculate the phase difference as a fraction of an oscillation.

    4) Convert to degrees or radians.

    18

  • Question 2 A body oscillating with SHM has a period of 1.5s and amplitude of 5cm. Calculate its frequency and maximum acceleration.

    (a) f = 1 / T = 1 / 1.5s frequency = 0.67 Hz

    (b) a = - (2Ο€f )2 x maximum acceleration is when x = A (the amplitude) a = - (2Ο€f )2 A

    = - (2Ο€ x 0.6667Hz)2 x 0.015m maximum acceleration = 0.88 ms-2

  • Question 3: Calculate the phase difference

    between the two oscillations in

    degrees and in radians.

    Answer:

    The time interval, 𝑑, is 17 ms.

    Period, 𝑇 = 60 ms.

    Phase difference,

    πœ™ = 𝑑

    𝑇 =

    17

    60 = 0.283 oscillation

    Phase difference in degree:

    πœ™ = 0.283 Γ— 360Β° = 102Β°

    Phase difference in radian:

    πœ™ = 0.283 Γ— 2πœ‹ = 1.78 rad

    20

  • Question 4

    The motion of a body in simple harmonic motion is described by the

    equation:

    π‘₯ = 4.0 cos(2πœ‹π‘‘ + πœ‹

    3 )

    Where π‘₯ is in metres, 𝑑 is in seconds and (2πœ‹π‘‘ + πœ‹

    3 ) is in radians.

    (a) What is (i) the displacement, (ii) the velocity, (iii) the acceleration

    and (iv) the phase when 𝑑 = 2.0 s? (b) Find (i) the frequency and (ii) the period of the motion.

    21

  • 22

    Solution:

    (a) (i) Displacement, π‘₯ = 4.0 cos(2πœ‹π‘‘ + πœ‹

    3 )

    when, 𝑑 = 2.0 s,

    π‘₯ = 4.0 cos(4πœ‹ + πœ‹

    3 )

    =2.0 m.

    (ii) Velocity, 𝑣 = 𝑑π‘₯

    𝑑𝑑 = βˆ’2πœ‹ 4.0 sin(2πœ‹π‘‘ + πœ‹/3)

    When 𝑑 = 2.0s,

    𝑣 = βˆ’8.0πœ‹ sin(4πœ‹ + πœ‹

    3 )

    = βˆ’21.8 ms-1

  • Solution (continued):

    (iii)Acceleration, π‘Ž = 𝑑𝑣

    𝑑𝑑 = βˆ’ 4πœ‹2 4.0 cos(2πœ‹π‘‘ +

    πœ‹

    3 )

    When 𝑑 = 2.0 s,

    π‘Ž = βˆ’ 4πœ‹2 4.0 cos 4πœ‹ + πœ‹

    3 = βˆ’79.0 ms-2.

    (iv) Phase = (2πœ‹π‘‘ + πœ‹

    3 )

    When t=2.0s, phase = (4πœ‹ + πœ‹

    3 ) rad = 13.6 rad.

    (b) (i) πœ” = 2πœ‹, 𝑓 = πœ”

    2πœ‹ =

    2πœ‹

    2πœ‹ = 1 Hz

    (ii) 𝑇 = 1

    𝑓 = 1s

    23

  • Simple Harmonic Motion

    Consider a simple harmonic oscillation, described by the formula

    π‘₯ = π‘₯0 cos(πœ”π‘‘)

    The velocity of its vibration is described as

    𝑣 = 𝑑π‘₯

    𝑑𝑑 = βˆ’πœ”π‘₯0 sinπœ”π‘‘

    The acceleration

    π‘Ž = 𝑑𝑣

    𝑑𝑑 = βˆ’πœ”2π‘₯0 cosπœ”π‘‘

    But π‘₯0 cosπœ”π‘‘ = π‘₯.

    Hence,

    π‘Ž = βˆ’πœ”2π‘₯.

    24

  • Acceleration in SHM

    25

    x

    a

    + A- A

    - Ο‰2 x

    + Ο‰2 x

    gradient = - Ο‰2 or - (2Ο€f )2

  • Simple Harmonic Motion

    𝑑2π‘₯

    𝑑𝑑2 = βˆ’πœ”2π‘₯

    𝑣 𝑑𝑣

    𝑑π‘₯ = βˆ’πœ”2π‘₯

    By separation of variables,

    𝑣𝑑𝑣 = βˆ’πœ”2π‘₯𝑑π‘₯ Integrating them,

    𝑣2

    2 = βˆ’

    πœ”2π‘₯2

    2 + 𝑐

    When π‘₯ = Β±π‘₯0, 𝑣 = 0.

    𝑐 = πœ”2π‘₯0

    2

    2 Hence, 𝑣2 = πœ”2 π‘₯0

    2 βˆ’ π‘₯2

    Or

    When v= max,

    𝑣 = Β±πœ” π‘₯0 2 βˆ’ π‘₯2

    26 𝑣0 = Β±πœ”π‘₯0

  • Simple Harmonic Motion

    𝑂

    A B

    -π‘₯ +π‘₯

    +π‘Ž βˆ’π‘Ž

    π‘₯ = βˆ’π‘₯0 𝑣 = 0 𝑑2π‘₯

    𝑑𝑑2 = πœ”2|π‘₯|

    Max. PE

    Zero KE

    π‘₯ = 0 𝑣 = Β±πœ”π‘Ž 𝑑2π‘₯

    𝑑𝑑2 = 0

    Zero PE

    Max KE

    π‘₯ = +π‘₯0 𝑣 = 0 𝑑2π‘₯

    𝑑𝑑2 = βˆ’πœ”2|π‘₯|

    Max. PE

    Zero KE

    27

  • Example:

    28

  • Answer: (a) Amplitude =0.15 m

    (b) Period 𝑇 = 2.0s

    (c) Frequency 𝑓 = 1

    𝑇 =

    1

    2.0 = 0.5 Hz

    (d) Angular velocity πœ” = 2πœ‹π‘“ = 3.14 rad sβˆ’1

    (e) (i) When π‘₯ = 0, 𝑑2π‘₯

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