IB DP Physics: SL复习笔记4.1.2 Simple Harmonic Oscillations

Simple Harmonic Oscillations

  • Simple harmonic motion (SHM) is defined as follows:

The motion of an object whose acceleration is directly proportional but opposite in direction to the object's displacement from a central equilibrium position

  • An object is said to perform simple harmonic oscillations when all of the following apply:
    • The oscillations are isochronous
    • There is a central equilibrium point
    • The object's displacement, velocity and acceleration change continuously
    • There is a restoring force always directed towards the equilibrium point
    • The magnitude of the restoring force is proportional to the displacement

Conditions for SHM

  • The defining conditions of simple harmonic oscillations are that the restoring force and the acceleration must always be:
    • Directed towards the equilibrium position, and hence, is always in the opposite direction to the displacement
    • Directly proportional to the displacement

a ∝ −x

  • Where:
    • a = acceleration (m s−2)
    • x = displacement (m)

The Restoring Force

  • One of the defining conditions of simple harmonic motion is the existence of a restoring force
  • Examples of restoring forces are:
    • The component of the weight of a pendulum's bob that is parallel and opposite to the displacement of the bob
    • The force of a spring, whose magnitude is given by Hooke's law

4.1.2-Restoring-Force-Examples

For a pendulum, the restoring force is provided by the component of the bob's weight that is perpendicular to the tension in the pendulum's string. For a mass-spring system, the restoring force is provided by the force of the spring.

  • For a mass-spring system in simple harmonic motion, the relationship between the restoring force and the displacement of the object can be written as follows:

F = – kx

  • Where:
    • F = restoring force (N)
    • k = spring constant (N m–1)
    • x = displacement from the equilibrium position (m)

4-1-2-restoring-force-graph_sl-physics-rn

Graph of force against displacement for an object oscillating with SHM

  • Force and displacement in SHM have a linear relationship where the gradient of the graph represents the constant
    • In this case, the spring constant k
  • An object in SHM will also have a restoring force to return it to its equilibrium position
    • This restoring force will be directly proportional, but in the opposite direction, to the displacement of the object from its equilibrium position

Acceleration & Displacement

  • According to Newton's Second Law, the net force on an object is directly proportional to the object's acceleration, F ∝ a for a constant mass

F = ma

  • Where
    • F = force (N)
    • m = mass (kg)
    • a = acceleration (m s−2)
  • Since F = ma (Newton's second law), and F = −kx (Hooke's law), the equations can be set as equal to one another:

ma = – kx

  • Rearranging to show the relationship between acceleration and displacement gives:
  • This equation shows that
    • There is a linear relationship between the acceleration of the object moving with simple harmonic motion and its displacement from its equilibrium position
  • The minus sign shows that when the mass on the spring is displaced to the right
    • The direction of the acceleration is to the left and vice versa
    • In other words, a and x are always in opposite directions to each other
  • This equation shows acceleration is directly proportional but in the opposite direction to displacement for an object in SHM

a ∝ −x

  • Therefore, it can be stated that:

a = −kx

  • Where
    • a = acceleration
    • k = is a constant but in this instance not the spring constant
    • x = displacement
  • Note that in physics, k is the standard letter used for an undefined constant

4-1-2-acceleration-graph_sl-physics-rn

Graph of acceleration against displacement for an object oscillating with SHM

Worked Example

A pendulum's bob oscillates about a central equilibrium position. The amplitude of the oscillations is 4.0 cm. The maximum value of the bob's acceleration is 2.0 m s–2.

Determine the magnitude of the bob's acceleration when the displacement from the equilibrium position is equal to 1.0 cm.

You may ignore energy losses.

Step 1: List the known quantities

    • Amplitude of the oscillations, x0 = 4.0 cm = 0.04 m
    • Maximum acceleration, a = 2.0 m s–2
    • Displacement, x = 1.0 cm = 0.01 m

Remember to convert the amplitude of the oscillations and the displacement from centimetres (cm) into metres (m)

Step 2: Recall the relationship between the maximum acceleration a and the displacement x

    • The maximum acceleration a occurs at the position of maximum displacement x = x0

a = – kx0

Step 3: Rearrange the above equation to calculate the constant of proportionality k

Conditions-for-Simple-Harmonic-Motion-Worked-Example-k-equation

Step 4: Substitute the numbers into the above equation

Conditions-for-Simple-Harmonic-Motion-Worked-Example-k-calculation

k = – 50 s–2

Step 5: Use this value of k to calculate the acceleration a' when the displacement is x = 0.01 m

a' = – kx

a' = – (– 50) s–2× 0.01 m

a' = 0.50 m s–2

 

 

 

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