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The concept of simple harmonic motion is best illustrated through a simple example.  Consider the system shown in the figure below.  A block of mass m rests on a frictionless surface, and one end of the block is attached to a spring, which in turn is attached to a fixed wall.

When the spring is neither stretched or compressed, there are no forces acting on the block and it remain stationary at the equilibrium position.  However, if the mass is pulled to the right, as shown in the figure, then the spring stretches and it exerts a restoring force on the mass.  If the spring is only stretched a small amount, the restoring force F is given by Hooke’s Law:

where k is known as the “spring constant” of the spring.  The minus sign arises because the force acts in the opposite direction to the displacement x.  Now Newton’s Second Law tells us that the force F acting on the block equals the mass m of the block times its acceleration.  Writing the acceleration of the block as the second derivative of the displacement x and using Newton’s Second Law yields the following equation of motion for the block:

This can be rewritten slightly into the following form:

This is the characteristic equation of simple harmonic motion.  It shows that the magnitude of the acceleration of the block is proportional to the displacement, and acts in the opposite direction to the displacement.  This is the required condition for simple harmonic motion, and hence the block in the diagram above undergoes simple harmonic motion.

Now, by using the following results from differential calculus,

it is easy to verify that the solution of the simple harmonic motion equation above is

To determine the constants A and B, it is necessary to consider the initial state of motion of the block.  The initial state of motion is characterised by the initial position x and velocity v of the block at time t = 0.  For arbitrary initial position and velocity, it is easy to show that

and hence the complete solution for the displacement x as a function of time is given by

Note that the amplitude of the motion is constant in time.  To illustrate simple harmonic motion, let us assume that the initial velocity is zero at time t = 0.  The solution above then simplifies to

This is illustrated in the following diagram, which shows the displacement x of the block as a function of time, for an initial displacement of unity.

At point A, the block is in its initial displaced position.  At point B the displacement is zero, but the gradient of the displacement curve attains its greatest negative value – the speed of the block is greatest therefore at point B.  At point C the block has reached the far opposite end of its motion.  The gradient at C is zero, so the velocity is zero and this is the point at which the block “turns around” to move in the opposite direction.  At point D the block once again attains its greatest speed, but in the opposite direction compared with point B.  At point E, the block has returned to its initial position.  This motion repeats itself indefinitely.

The “time period” T of the harmonic motion is the time required for the oscillations to complete one cycle, that is to move from A to E in the above figure.  Thus

The previous discussion is illustrated in the following figure, which shows the motion of the block at the points A through to E.  The arrows represent the direction of motion of the block.  At points C and E, the block arrives from one direction and then turns to move in the opposite direction.

In a later section we will look at a more interesting system, the so-called simple pendulum.  This consists basically of a mass attached to a string that is free to swing in a vertical plane, rather like the pendulum of a clock.  Such a system also exhibits simple harmonic motion when the angle of swing is small.

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