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## Simple Harmonic Motion II, Physics tutorial

:Mass Hanging from a Coiled Spring:Period of OscillationFrom Hooke's law, extension of the coiled spring is directly proportional to force causing it. In the diagram below, the mass hanging from coiled spring to exert the downward tension mg on spring. This is exactly what occurs.

Let extension produced by downward tension be l, and if k is tension needed to produce the unit length of spring than stretching tension is also kl. (k is also called as spring constant and is measured in Nm-1). This signifies that,

mg = kl

When we now pull down mass below the equilibrium position as shown, a distance x, stretching tension becomes k(l + x). This is the same as tension in spring acting upwards as given in Figure. Therefore we can represent resultant restoring force upwards on mass as K(l + x) - mg

= Kl + kx - mg

but mg = kl

The resultant restoring force = kx

When release the mass after extension it begins moving up and down continuously which we call oscillatory motion. If at the extension x it has acceleration a, then its equation of motion will be

ma = -kx

Minus sign illustrates that at instant while displacement x is downwards (i.e. positive) acceleration a, is directed upwards equilibrium position (i.e. negative).

Therefore a = -k/mx = ω

^{2}xWhat ω2 = k/m. As m and k are positive constants we see that ω

^{2}also is a positive constant. As a result acceleration a is constant and this is condition for the motion to be simple harmonic. We hence conclude that motion of mass is simple harmonic as long as Hooke's law is obeyed.Period T is given by

T = 2Π/ω = 2Π√m/k

Squaring both sides we have that

T

^{2}= 4Πm/kIf in the experiment, we differ mass m and record square of corresponding periodic time, T on plotting graph of T versus m, a straight line graph will be expected. This kind of experiment has really been done several times over. It was seen that straight line graph didn't pass through origin. And clarification was sought by scientists and it was found that it was because mass of spring itself was not taken in consideration. So it was necessary to find out effective mass undergoing simple harmonic motion.

:Simple PendulumSimple pendulums are at times utilized as the example of simple harmonic motion, SHM, as their motion is periodic. They also fit criteria that bob's velocity is maximum as it passes through equilibrium and acceleration is minimal while at each endpoint. But the deeper understanding of behavior of bob will demonstrate us that pendulums don't truly fit SHM model.

When bob is displaced vertically to point A through extremely small angle Θ as given in diagram and then released, it oscillates to and fro, in the vertical plane, about equilibrium position 0. The motion of bob is seen to trace the arc of circle with radius l (suppose bob is a point man). We shall see that motion is simple harmonic about O.

Now, consider arc thread by the bob be OA = x and angle of displacement OBA = Θ at some instant of time when bob is at point A. At that moment, forces on the bob are the weight of bob mg acting vertically downwards as illustrated and P the tension in string (or thread). But mg has tangential component mgSinΘ which serves as balancing restoring force towards O and radial component mgCosΘ balancing tension P in the string. If a is acceleration of the bob along the arc at A because of mgsinΘ then from Newton's law of motion we have,

ma = --- mgsin Θ

Displacement x is measured from 0 towards A. along arc OA while negative sign illustrates that restoring force is applying opposite to direction of displacement i.e. towards Θ. For very small angle Θ, mathematics allows us to suppose that sin Θ = Θ in radians (for instance, if Θ = 5

^{o}, Sin Θ = 0.0872 and Θ = 0.0873 rad.) and x = lΘ. Therefore Θ = x/lma = mgΘ = -mg(x/l)

Therefore a = -(g/l)x

Setting g/l = ω

^{2}We have

a = -ω

^{2}xWe can then compute that motion of bob is simple harmonic if oscillations are of small amplitude Θ as we assumed. In short Θ must not exceed 100. Period T for simple pendulum, is provided by

T = 2Π/w = 2Π√g/l

2Π√l/g

We notice that T doesn't depend on amplitude of oscillations. For the specific location on surface of earth where g is constant, period of oscillation of the simple pendulum is seen to rely only on length of pendulum.

:Energy of Simple Harmonic MotionDuring simple harmonic motion of an object, there is a constant interchange of energy of the object between its kinetic and potential forms. Note that if there is no influence of resistive forces (i.e. damping forces) on the object, its total energy E = (K. E. + P. E) is constant.

Kinetic Energy, K. E:The velocity of a particle N of mass m at a distance x from ist centre of oscillation O is given by:

v = +w√r2 - x2

The kinetic energy K. E. at x, say, is

K.E. = 1/2mv

^{2}= 1/2mw^{2}(r^{2}- x^{2})The Potential Energy, P. E:During the motion of the particle N from O towards A or B, work is done against the force trying to restore it to O. Therefore, the particle loses some K. E. but gains some P. E. When x = 0, the restoring force is zero. But at any displacement, say, x the force is mω2x because the acceleration at that point has magnitude ω2x.

Thus, average force on N while moving to displacement x

= (0+mω

^{2}x)/2 = 1/2mω^{2}xTherefore work done = average force x displacement in the direction of force

= (1/2mω

^{2}x).(x)= 1/2mω

^{2}x^{2}Therefore P.E. at displacement x = 1/2mω

^{2}x^{2}Total Energy, E:The total energy at displacement x is then given by K. E + P. E

Total Energy E = 1/2mw

^{2}(r^{2}- x^{2}) + 1/2mω^{2}x^{2}= 1/2mω

^{2}x^{2}We see that this value is constant and does not depend on x. It is also directly proportional to the product of (i) mass (ii) the square of the frequency (iii) the square of the amplitude.

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