We have seen that for a system to execute simple harmonic motion, i»must have two parts: one which can store potential energy (like spring) and the other capable of storing kinetic energy (such as mass). We will now study physical systems executing SHM using techniques developed for our model spring-mass system.

Simple Pendulum:

A simple pendulum is an idealized system consisting of a point mass (bob) suspended by an inextensible, weightless string. As the bob of mass m is displaced by an angle θ from its equilibrium position, the restoring force is provided by the tangential component of the weight mg along the arc. It is given by

F = -mgsinθ

Equation of motion of bob is, hence,

md²x/dt² = -mgsinΘ..............................................Eq.1

Bob is moving along arc whose length at any instant is provided by x. If corresponding angular displacement from equilibrium position is Θ, then length of arc is

x = lΘ..............................................Eq.2

Where l is length of string by which bob is suspended. Differentiating Eq.2 twice with respect to t, and substituting result in Eq.1, we get

d²Θ/dt² = -(g/l) sin Θ..............................................Eq.3

For small angular displacements, sin Θ may be approximated to Θ. In this approximation, Eq.3 takes form

d²Θ/dt² + ω₀²Θ = 0..............................................Eq.4

Where ω₀ = √g/l

Eq.4 is exactly of standard form illustrating that pendulum executes simple harmonic motion. Time period of oscillation is provided by

T = 2π/ω₀ = 2π√l/g..............................................Eq.4

By analogy, write general solution of the Eq.4 as

Θ = Θ_m cos(ω₀t + Φ)

When amplitude of oscillation is not small, solve general Eq.3. Time period that can be stated in form of a series involving maximum angular displacement Θ_m is provided by

T = 2π√l/g(1 + (1/2²)sin²Θ_m/2 + (1/2²)(3²/4²)sin⁴Θ_m/2+.....)

Compound Pendulum:

The compound pendulum is the rigid body able to oscillate freely about horizontal axis passing through it. At equilibrium position, the centre of gravity G lies vertically below point of suspension S. Let distance SG be l . If pendulum is given a small angular displacement Θ at any moment, it oscillates over same path. The restoring torque about S is -mgl sinΘ and it tends to bring pendulum towards equilibrium position. If I is moment of inertia of body about horizontal axis passing through S, restoring torque equals Id²Θ/dt². Therefore, the equation of motion can be written as

Id²Θ/dt² = -mglsinΘ..............................................Eq.5

For small angular displacement, sin Θ ≈ Θ and Equation 5 takes form

d²Θ/dt² + mgl/IΘ = 0..............................................Eq.6

This equation illustrates that compound pendulum executes SHM and time period is provided by

T = 2π/ω₀ = 2π√I/mgl..............................................Eq.7

There is very helpful and significant theorem of parallel axes in study of moment of inertia. According to the theorem, moment of inertia I of the body about any axis and inertia Ig about parallel axis passing through center of gravity are connected by relation

I = Ig + ml²..............................................Eq.8

Where l is the distance between the two axes and Ig = mk_r². Quantity k_r is radius of gyration of body about axis passing through G. It is radial distance at which whole mass of the body could be placed without any change in moment of inertia of body about that axis.

On substituting the expression for I from Eq.8 in Eq.7 we obtain

 T = 2π√(k²_r + l²)/gl..............................................Eq.7

On comparing expression for T with that given by Eq.4 for simple pendulum, two periods become equal if l in Eq.4 is replaced by L = √(k²r/l)+l . This is known as length of the equivalent simple pendulum. If we produce line SG and take point O on it such that SO = k²r/l + l, then O is known as centre of oscillation.

Torsional Systems:

If one end of the long thin wire is clamped to the rigid support and other end is fixed to centre of massive body like disc, cylinder, sphere or rod, then arrangement is known as torsional pendulum. It is generally utilized to find out moment of inertia of regular and irregular bodies. Ammeters, voltmeters and moving coil galvanometers are other estimating devices where restoring torque is given by spiral springs or suspension fibres.

When the torsional system is twisted and then left free, it executes torsional oscillations in the horizontal plane. For the angular displacement Θ the restoring torque is Θ. Here, k_t is constant that depends on properties of wire. If I is moment of inertia of system about axis of rotation and d²Θ/dt² is angular acceleration, then equation of motion is

Id²Θ/dt² = -k_tΘ

or d²Θ/dt² = ω₀²Θ..............................................Eq.8

Where ω₀ = √k_t/l. This is exactly of standard form. Therefore, motion is SHM and period of oscillation is

T = 2π√I/k_t..............................................Eq.8(a)

This expression for T has no approximation. This means that time period for large amplitude oscillations will also remain same, provided the elastic limit of suspension wire is not exceeded.

An L-C Circuit:

L-C circuit has no moving parts, but electric and magnetic energies in circuit play roles analogous to potential and kinetic energies respectively for the spring-mass system. For ease, we suppose that inductor has no resistance.

In the pendulum, mean position is taken as equilibrium state. What is the equilibrium state in an L-C circuit? It corresponds to state when there is no current in circuit. It may be disturbed by charging or discharging capacitor. Let capacitor be given the charge Q₀ coulomb. Then voltage across capacitor plates will be Q_o/C. Now if circuit is disconnected, capacitor discharges through inductor. Consequently current begins building up in circuit slowly and charge on plates of capacitor decreases. At any time t, let current in circuit be q and charge on capacitor plates be q. Then voltage drop across inductor will be

V_L = -Ldl/dt

This should be equal to voltage V_c = q/C across capacitor plates at that time. Therefore, we can write

V_c = V_L

or q/C = -LdI/dt..............................................Eq.9

As = I = dq/dt and dI/dt and dI/dt = d²q/dt², Eq takes the form

d²q/dt² + ω₀²q = 0..............................................Eq.10

Where ω₀² = 1/LC. This signifies that one can have wide range of frequencies by changing values of L and C. That is how you tune different stations in radio sets.

Equation represents SHM and has solution

q = q₀cos(ω₀t + Φ) ..............................................Eq.11

This illustrates that charge oscillates harmonically with period

T = 2π√LC..............................................Eq.12

Differentiating Eq.11 with respect to time, we get instantaneous current

I = Q₀ω₀sin(ω₀t + Φ)

= I₀cos(ω₀t + Φ + π/2)

Where I₀ = Q₀ω₀

Therefore current leads charge in phase by π/2.

Now compute energy stored in inductor L and the capacitor C at any instant t. As the current increases from zero to I in time t, energy stored in the inductor, EL, is attained by integrating instantaneous power with respect to time, i.e.

E_L = -∫₀^tIV_Ldt

Negative sign means that work is done against, rather than by emf. On substituting for VL, we get

EL = L∫₀^I(dI/dt)Idt = 1/2LI²

The energy stored in capacitor at time t is

Ec = q²/2c

Therefore total energy

E = E_L + E_C = 1/2LI² + 1/2q²/C..............................................Eq.13

This expression, for total energy is like one for mechanical oscillator (E = 1/2mv² + 1/2kx²).As q and I differ with time, inductor and capacitor exchange energy periodically. This is like the energy exchange in spring-mass system. Further, mass and inductor play analogous roles in mechanical and electrical systems, respectively.

Diatomic Molecule: Two-Body Oscillations:

The diatomic molecule like HCl is the example of two-body system that can oscillate along the line joining two atoms. Atoms of the diatomic molecule are coupled through forces that have electrostatic origin. Bonding between them may be likened to the spring. Therefore we may consider the diatomic molecule as the system of two masses joined by the spring. Consider oscillations of such a system. Assume that two masses m₁ and m₂ are joined by the spring of force constant k. Masses are constrained to oscillate along axis of the spring. Let r₀ be normal length of spring. We select X -axis along line joining two masses. If X₁ and X₂ are coordinates of two ends of spring at time t, change in length is given by

x = (X₂ - X₁) - r₀..............................................Eq.14

For x > 0, x = 0 and x < 0, spring is extended, normal and compressed respectively. Assume that at a given instant of time spring is extended, i.e. x > 0. Though spring applies same force (kx) on two masses, force F₁ (= kx) acting on m₁ opposes force F₂(=-kx) on m₂, i.e.,

F₁ = kx and F₂ = -kx

According to Newton's second law, above equation can be written as: m₁(d²X₁/dt²) = kx and m₂(d²x₂/dt²­) = -kx

On rearranging terms, we get

d²X₁/dt² = kx/m₁..............................................Eq.15(a)

and

d²X₁/dt² = kx/m₁..............................................Eq.15(b)

On subtracting one from the other, we get

d²(X₂ - X₁)/dt² = -(1/m₁ + 1/m₂)kx

Since r₀ signifies the constant length of spring, Eq.14 tells us that

d²x/dt² + k/μx = 0..............................................Eq.16

Where μ(1/m₁ + 1/m₂)-1 = m₁m₂/m₁+m₂is called the reduced mass of the system.

Eq.16 explains simple harmonic oscillation of frequency

v₀ = 1/2π√k/μ

This means that the diatomic molecule acts as single object of mass μ, joined by the spring of force constant k.

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