Introduction:

Each hand of the clock comes back to given position after lapse of the certain time. This is a well-known example of periodic motion. When the body in periodic motion moves to and- fro (or back and forth) about its position, motion is known as vibratory or oscillatory.  Familiar examples of oscillatory motion are:

The oscillating bob of the pendulum clock, piston of engine, vibrating strings of the musical instrument, oscillating uranium nucleus before it fissions, even large scale buildings and bridges may sometimes suffer oscillatory motion. Several stars show periodic variations in brightness). Such oscillations, left to themselves, don't continue indefinitely, i.e., they slowly die down because of different damping factors like friction and air resistance, etc.

Oscillations of a Spring-mass System:

The spring-mass system comprises of the spring of negligible mass whose one end is fixed to the rigid support and other end carries the block of mass m that lies flat on horizontal frictionless table figure(a). Let us take x -axis to be along length of the spring. When mass is at rest, mark the point on it and we state origin of axis by this point. That is, at equilibrium mark lies at x = 0.

If spring is stretched by pulling mass longitudinally, because of elasticity the restoring force comes into play that tends to bring mass back towards equilibrium position in figure (b). If spring were compressed restoring force would tend to extend spring and restore mass to equilibrium position as shown figure (c).

The more you stretch/compress the spring, the more will be the restoring force. So the direction of the restoring force is always opposite to the displacement. If total change in the length is small compared to the original length, then the magnitude of restoring force is linearly proportional to the displacement. Mathematically, we can write

F = -kx......................................Eq.1

The negative sign means that restoring force opposes displacement. Quantity k is known as spring constant or force constant of spring. It is numerically equal to magnitude of restoring force applied by spring for unit extension. Its SI unit is Nm^-1.

Differential equation of simple harmonic motion:

Now determine differential equation that explains oscillatory motion of the spring-mass system. Equation of motion of such a system is provided by equating two forces acting on mass:

Mass X acceleration = restoring force Or md²x/dt² = -kx

Where d²x/dt² is acceleration of the body.

In this equation, equilibrium position of body is taken as origin, x = 0.

You will note that quantity k/m has units of Nm^-1kg^-1 = (kg.mg^-2)kg^-1m^-1 = s^-2. Therefore we can replace k/m by ω₀² where ω₀ is known as angular frequency of oscillatory motion.

Then above equation takes form

md²x/dt² + ω₀ = 0......................................Eq.2

It may be remarked here that Eq.2 is differential form of Eq.1 and explains simple harmonic motion in one dimension.

The differential equation containing terms involving only first power of variable and its derivatives is called as the linear differential equation. If such equation has no term independent of variable it is said to be homogeneous. Therefore Eq. 2 is the second order linear homogeneous equation. Its solution will have two arbitrary constants.

Solution of differential equation for shm:

To determine displacement of mass at any time t, solve Eq. 2 subject to given initial conditions. The close inspection of this equation illustrates that x must be such a function that second derivative with respect to time is negative of function itself, except for the multiplying factor o>o.

A general solution for x (t) can therefore be stated as linear combination of both sine and cosine terms, i.e.,

x(t) A₁ cos αt + A₂ sin αt......................................Eq.3

Putting A₁ = AcosΦ and A₂sinΦ, we get

x(t) = A cos (αt + Φ)

Differentiating equation twice with respect to time and comparing resultant expression with Eq. 2 we get α = ±ω₀. Negative sign is dropped as it provides negative frequency that is physically absurd quantity.

Substituting α = ω₀ in above equation, we get

x(t) = Acos(ω₀t + Φ). ......................................Eq.4

Constant A and Φ in Eq.4 are determined by initial conditions on displacement (x) and velocity dx/dt.

Let us suppose that mass is held steady at some distance a from equilibrium position and then released at t = 0. Therefore initial conditions are: at t = 0, x = a and dx/dt = 0.

Then, from Eq.4 we would have: x (at t = 0) = AcosΘ = a and dx/dt(at t= 0) = -Aω₀sinΦ = 0

The second condition tells us that Θ is either zero or nπ (n = 1, 2,...). We reject second option as first condition requires cos Φ to be positive. Therefore with above initial conditions, Eq.4 has simple form

x = acosωt......................................Eq.5

Phase and Amplitude:

Quantity (ω₀t + Φ), in Eq.4 is known as phase angle or phase angle of system at t = 0, also known as initial phase we start estimating displacement. If at x = x₀, then from Eq.4 it follows that

x₀ = acosΦ

We know that value of sine and cosine functions is between 1 and -1. When cos(ω₀t + Φ) = 1 or -1, displacement has maximum value. Let us indicate it by a or -a. Quantity a is known as amplitude of oscillation.

Thus, rewrite Eq.4 as

x(t) = cos(ω₀t + Φ) ......................................Eq.6

Time Period and Frequency:

If we put t = t + (2π/ω₀) in Eq.6, we get

x(t) = acos[ω0(t + 2π/ω₀) + Φ]

= a cos[ω₀t + 2π + Φ]

= a cos[ω₀t + Φ]

That is, displacement of particle repeats itself after interval of time 2π/ω₀. In other words, oscillating particle completes one vibration in time 2π/ω₀. This time is known as period of vibration or time period. We denote it by T:

T = 2π/ω₀......................................Eq.7

For spring-mass system ω₀² = k/m, so that

T = 2π√m/k......................................Eq.8

Number of vibrations executed by oscillator per second is known as frequency. Unit of frequency is Hertz (Hz). Denoting it v₀, we have for spring-mass system

v₀ = 1/T = 1/2π√k/m......................................Eq.9

This signifies that stiffer the spring, higher will be the frequency of vibration.

Velocity and Acceleration:

We know that displacement of the mass executing simple harmonic motion is provided by

x = acos(ω₀t + Φ)

Thus, instantaneous velocity, which is time derivative of displacement, is provided by

v = dx/dt = -ω₀asin(ω₀t + Φ)....................................Eq.10

We can rewrite it as: v = ω₀acos(π/2 + ω₀t + Φ)....................................Eq.11

The value of v at any point v. To this end, rewrite Eq.10 as

v = -ω₀[a² - a²cos(ω₀t + Φ)]1/2 for -a ≤ x ≤ a....................................Eq.12

Acceleration is first time derivative of velocity. From Eq.10

dv/dt = -ω₀²acos(ω₀t + Φ)

= ω₀²acos(π + ω₀t + Φ) ....................................Eq.13

Obviously, in terms of displacement,

dv/dt = -ω₀²x....................................Eq.14

ω₀a is velocity amplitude and ω₀²a is acceleration amplitude, and velocity is ahead of displacement by π/2 and acceleration is ahead of velocity by π/2.

Transformation of energy in oscillating systems: potential and kinetic energies:

In spring-mass system when mass is pulled, spring is elongated. Amount of energy needed to lengthen spring through the distance dx is equal to work done in bringing about this change. It is given by dW - dU = F₀ dx where F₀ is applied force. This force is balanced by restoring force. That is, magnitude is same as that of F and we can write F₀ = kx. Thus, energy needed to lengthen spring through the distance x is

 U = ∫₀^xF₀dx = k∫₀^xxdx = 1/2kx²....................................Eq.15

This energy is stored in spring in form of potential energy and is liable for oscillations of spring-mass system. On substituting for the displacement from Eq.6 in Eq. 15

u = 1/2ka²cos(ω₀t + Φ) ....................................Eq.16

Note that at t = 0, potential energy is

U₀ = 1/2ka²cos²Φ....................................Eq.17

As mass is released, it moves towards equilibrium position and potential energy begins changing in kinetic energy (K.E). Kinetic energy at any time is given by K.E.= 1/2mv² .

Using Eq.10, we get

K.E. = 1/2mω₀²sin2(ω₀t + Φ)

= 1/2ka²sin²(ω₀t + Φ) ....................................Eq.18

As ω₀² = k/m

One can also state K.E. in terms of displacement by writing K.E. in terms of displacement by writing

K.E. = 1/2ka²[1 - cos²(ω₀t + Φ)]

= 1/2ka² - 1/2ka²cos²(ω₀t + Φ) ....................................Eq.19

= 1/2ka² - 1/2kx² = 1/2k(a² -x²) ....................................Eq.20

This illustrates that when oscillating body passes through equilibrium position (x = 0), kinetic energy is maximum and equal to 1/2ka².

Calculation of average values of quantities associated with shm:

For any complete cycle in every case, area under curve for first half is accurately equal to area under the curve in the second half and the two are opposite in sign. Therefore over one complete cycle algebraic sum of these areas is zero. This signifies that average values of displacement, velocity and acceleration over one complete cycle are zero. Time average of kinetic energy over one complete cycle is stated as

<K.E.> = (∫₀^TK.Edt)/T....................................Eq.21(a)

On substituting for K.E from Eq. (1.19), we get

<K.E.> = ka²/2T∫₀^Tsin²(ω₀t + ω)dt....................................Eq.21(b)

On solving the integral in Eq.21(b). find that value is T/2. So, the expression for average kinetic energy reduces to

<K.E.> = ka²/4....................................Eq.22

Similarly, one can show that the average value of potential energy over one cycle is

<U> = ka2/4 ....................................Eq.23

That is, the average kinetic energy of the harmonic oscillator is equal to average potential energy over one complete period. Therefore sum of average kinetic and average potential energies is equal to total energy :

<K.E.> + <U> = 1/4ka² + 1/4ka²

= 1/2ka² = E

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