How do we Analyse Simple Harmonic Motion?

Many engineering systems consist of components which oscillate or vibrate in the presence of appropriate forces. This may be desirable or undesirable, or simply unavoidable. A theoretical understanding of the nature of vibration allows us to analyse and solve a wide range of engineering problems. In fact, many of the systems, including engines, presses, unbalanced rotors, and machines with belt drives, may all experience vibration in operation.

Any system which consists of a mass which rests on or is supported by a spring or otherwise flexible structure (and most structures have some flexibility) falls into this category. One of the most important factors is identifying the natural frequency of the system, which is the frequency at which resonance occurs. This is the frequency at which the system will naturally vibrate in the absence of any outside force.

Vibration can be categorised into free vibration, where there is no force generating the motion, and forced vibration, where a periodic force, such as an unbalanced rotating object, acts on the system. 

Vibration can also be categorised as damped or undamped. Damping is a force which causes vibrations to die away, and is present to some extent in most systems. We will be looking at undamped free vibration initially, and we will also briefly consider undamped forced vibration when we examine the phenomenon of resonance.

Notation

In vibration analysis we are dealing with displacement, velocity and acceleration and so it is common to use a short-hand notation as follows:

x = displacement 

$\stackrel{.}{x}$ = velocity = ^dx / _dt

$\ddot{x}$ = acceleration ^d2x / _dt²

where the over-dot represents differentiation with respect to time, t.

Free Response of Mass Spring System

The spring-mass system is a very basic system but its behaviour is typical of many systems found in engineering, and an understanding of its behaviour; simple harmonic motion and resonance, provides a great deal of information about a range of engineering systems.

In the derivation below, we will consider the free response of the system. That is, we consider a system where the mass is displaced slightly from its equilibrium position, and then released. There are no external forces acting on the mass other than the spring force.

A spring mass system shown at its equilibrium position. The corresponding position, velocity and acceleration are shown – position and velocity are towards the right, and are positive, and acceleration is negative since the mass is decelerating.

Consider a mass attached to a spring as shown above. We will assume that there are no friction or damping forces acting. If we displace the mass from its equilibrium position a distance x, it will feel a force from the spring equal to kx directed towards its original position. If the mass is released, this force will accelerate the mass towards the original position with a force which reduces to zero. As the mass passes its original position, the spring force begins to increase in the other direction, decelerating the mass.

The equation of motion of the mass can be described as below, assuming positive is to the right. If the mass is displaced to the right, it will feel a spring force acting to the left. Therefore, we can write the equation as:

m$\ddot{x}$ = -kx

Dividing across by m, we obtain an expression for acceleration

$\ddot{x}$ = –^k/_m x

This equation allows us to make two important observations about the motion of the mass:

The acceleration is always proportional to the displacement from the equilibrium position.

The acceleration is always directed towards the equilibrium position ( when the displacement is positive , acceleration is negative)

To solve this equation, we note that the acceleration, which is the second derivative of position, is always equal to minus a constant (^k/_m) multiplied by position (x). Therefore, any mathematical function which also has this property, is a candidate. In fact, trigonometric functions have this property, so we can try the following:

Suppose we assume that

x(t) = A cos (ωt)

where A is the amplitude (m) of  = A cos (ωt) the response and t is time (s) and  ω the frequency (rad/s) then if we differentiate with respect to time:

$\stackrel{.}{x}$ = – ωA sin (ωt)

$\ddot{x}$ =- ω² A cos (ωt) = -ω²x

Therefore:

 $\stackrel{}{\mathrm{x¨}}$ = – ω²x

where ω² = ^k/_m

The above solution will work for our equation of motion, so we can say that a spring-mass system which has been given an initial disturbance moves in a sinusoidal manner. If we plot the position of the mass against time, it has a sinusoidal path, and the maximum value of this displacement is A, so that the mass moves between maximum displacements of –A and +A from its rest position.

Natural Frequency

Let us look more closely at the motion of the mass. If given an initial disturbance, the mass will oscillate about its equilibrium position at a particular frequency, and in the absence of damping, will continue to do so at the same amplitude. The equation of motion is 𝑥(𝑡) = 𝐴 cos 𝜔𝑡. The quantity A is the amplitude of the motion, that is, the distance of the maximum displacement from the equilibrium position. Looking at the relationship between acceleration and position we have

 $\ddot{x}$ = -ω²x 

which gives us an expression for  ω as follows:

ω = √^k/_m

The quantity ω represents the angular frequency of the motion, and ranges from 0 to 2π over one complete oscillation. This represents the natural angular frequency of vibration (in radians per second), and is usually referred to as ωn. If we divide this quantity by 2π, we obtain fn ,the natural frequency of vibration in Hertz (cycles per second), and the inverse of this is the periodic time, which is the time in seconds for one oscillation.

ω_n  = √^k/_m   rad/s 

f_n = ¹/_2π  √^k/_m Hz 

Note T_p = ¹ / _fn

Where k = spring stiffness, f_n = natural frequency, T_p = periodic time ( s)

m = mass attached to the spring, ω_n = natural frequency in rad/s

Alternative expression

Another form of the simple harmonic motion equation can be obtained by noting the relationship between acceleration and velocity:

Since   $\stackrel{}{\mathrm{x¨}}$ = -ω²x  we can write

 ω = √^$\ddot{x}$/_x   = √^acceleration / _displacement

This equation is valid at any time and is valid for any system which executes simple harmonic motion.

Example

A mass of 5 kg is suspended from a spring of stiffness 2 kN/m. 

(i) Calculate the natural frequency of the spring. 

(ii) If the spring is initially displaced by 25 mm and released, calculate the maximum velocity and acceleration of the mass as it oscillates.

Note that the spring is initially in equilibrium, so we do not need to consider the weight of the mass as the spring has already deflected to account for this.

Solution:

(i) Natural frequency (and periodic time):

ω_n  = √^k/_m  = √²⁰⁰⁰ /₅ = 20 rad/s

f_n = ^ω_n / _2π = 3.18Hz

T = ¹/_fn = 0.31s

(ii) Maximum velocity and acceleration:

The formula for velocity is 

$\stackrel{}{\mathrm{x.}}$ = – ωx sin (ωt)

where A is the amplitude of the position. Since we initially displaced the mass by 25 mm before releasing it, A must be 25 mm. Since the maximum value of sin(ωt) is ±1, then the maximum absolute value of velocity must be ωA:

丨$\stackrel{.}{x}$丨_max = -ωA  = (20)(0.025) = 0.5m/s

$\ddot{x}$ =- ω² A cos (ωt) = -ω²x

A similar situation exists for acceleration, whose formula is $\ddot{x}$=- ω² A cos (ωt)

 Again, since the maximum value of the cosine term is ±1, we have a maximum value of acceleration of:

丨$\ddot{x}$丨_max = ω²A  = (20)²(0.025) = 10m/s² 

A plot of position, velocity and acceleration versus time for the above system is shown below. Note the relationship between each in terms of phase and magnitude. 

In particular note for time zero (when the mass is released from its displacement), the position is at its maximum, the velocity is zero and the acceleration has its maximum absolute value and is of the opposite sign to the position.

Plot of position, velocity and acceleration vs time for the above problem.

Calculating velocity and acceleration at any position:

We can use the equations for position, velocity and acceleration to calculate these values at any point in the oscillation rather than just the extreme points. 

Example 1:  

Calculate the velocity and acceleration when the particle reaches a displacement of 10 mm from equilibrium i.e. x = 0.01 m. We use the above values of A = 0.025 m, and ω = 20 rad/s.

x = A cos ωt

Let us rearrange this for time, t:

 t = ¹/_ω cos^-1( ^x/_A)

For x = 0.01m, we have (remember to use radians):

 t = ¹/₂₀ cos^{-1 ( 0.01/_0.025)}

t = 0.058s

Now we can calculate the velocity and acceleration at this time:

$\stackrel{.}{x}$ = -A ω sin ωt

$\stackrel{}{\mathrm{x.}}$= – (0.025)(20) sin (20 x 0.058) = -0.458m/s

$\stackrel{}{\mathrm{x¨}}$ = -A ω² cos ωt = – (0.025)(20²) cos( 20 x 0.058) = -4.011m/s²

Example 2:

A 2 kg block is placed on a frictionless surface. A spring with a force constant of k = 32 N/m is attached to the block, and the opposite end of the spring is attached to the wall. The spring can be compressed or extended. The equilibrium position is marked as x = 0 m.

Work is done on the block, pulling it out to x = +0.02 m. The block is released from rest and oscillates between x = +0.02 m and x = −0.02 m. The period of the motion is 1.57 s. Determine the equations of motion.

Solution

The angular frequency can be found and used to find the maximum velocity and maximum acceleration:

f_n = ¹/_T = ¹/_1.57 = 0.64Hz

ω = 2πf = 2π(0.64) = 4 rad/s

Maximum velocity:

丨$\stackrel{.}{x}$丨_max = ωx  = (4)(0.02) = 0.08m/s

Maximum acceleration:

丨$\ddot{x}$丨_max = ω²x  = (4)²(0.02) = 0.32 m/s²

$\ddot{x}$ = -A ω² cos ωt = – ω²x

All that is left is to fill in the equations of motion:

x(t) = A cos ωt = 0.02cos 4t

$\stackrel{.}{x}$ = -A ω sin ωt = – (4)(0.02) sin 4t = -0.08sin 4t

$\ddot{x}$ = -A ω² cos ωt = – (0.02)(4²) cos 4t = 0.32cos 4t

Significance

The position, velocity, and acceleration can be found for any time. It is important to remember that when using these equations, your calculator must be in radians mode

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