1

Simple harmonic oscillator

Vibrations are everywhere, including bridges, light, optics, music, crystal lattices, fundamental particles and so on.

Lets look and think about the motion of a mass on a spring. 1. The equilibrium position (e.p.) exists. 2. We draw the mass away from the equilibrium position and release it. 3. The mass starts from rest and moves towards the equilibrium position. 4. The mass has inertia, and so over-shoots the equilibrium position. When it passes through the equilibrium position it has the largest velocity in the cycle. 5. It reaches rest on the other side, starts to move back towards the equilibrium position. 6. The cycle repeats. The motion is periodic. So we expect the motion can be described by a perioic function. We know that sine and cosine are a periodic function, with a periodicity of 2. This means if we want the system to repeat in a time T , called the period, we might think the motion could be described by the functional form x(t) cos( 2 t). T (1)

This has the right form, as when t goes from 0 to T , the argument of the cosine goes from 0 to 2. So consider our mass on a spring. We have a spring of mass m, and label the distance from the equilibrium point as x. At this stage we assume no friction, and a massless spring. As we know from observing the mass on a spring, the motion is periodic, with a time for one cycle as the period T . The maximum amplitude is A, and if we start at the maximum amplitude we can write the initial conditions as x(0) x (0) = A = 0. (2)

We also dene the frequency as = 1/T , with units of Hz=/s, and the angular frequency as = 2. (3) The units of are rad/s, and it tells us how many radians the oscillator advances per unit time. For small displacements we assume the restoring force follows Hookes law, and we can write Frestore = kx (4)

with the minus sign giving us a force directed towards the equilibrium point. The constant k is called the spring constant. Using Newtons second law we get Frestore d2 x m 2 dt = kx = ma = kx (5)

and we are arrive at the key dierential equation d2 x = 2 x, dt2 (6)

where we write 2 = k/m. This is our general equation of simple harmonic motion. It is a linear, 2nd order, ordinary dierential equation. To solve it we appeal to our physical intuition and guess a trial solution which should obey the dierential equation (a function whose 2nd derivitive with respect to time gives the same function back, and is periodic). Our guess, and its derivitives are x(t) = A cos t, dx(t) = A sin t, dt d2 x(t) = A 2 cos t, dt2

(7)

which we can see obeys the dierential equation. In general we can have an initial arbitrary phase, giving us a second constant and the solution x(t) = A cos(t + ). (8)

Note that we can use a sine form of the solution simply by changing by /2. Recall sin(t) = cos(t /2). (9) The constants A and are determined by the initial conditions, whilst the angular frequency is given by the spring constant and the mass ( 2 = k/m), and so is a property of the system. We can also write the general solution as a sum of a sine and a cosine wave x(t) = a cos t + b sin t. (10)

The relationship between the two sets of constants is explored in the problem sheet. Finally the velocity is given by v(t) = dx = A sin(t + ), dt (11)

which is /2 out of phase with the position, with the velocity leading the position (the velocity peaks /2 before the position). Now we turn to the use of energy to study a simple harmonic oscillator. Energy considerations are a powerful tool, as energy is a scalar quantity and conservation laws apply. The kinetic energy is simply K = 1 mv 2 . The potential energy 2

of a spring is given by force times distance, so we write the work done going from x to x + dx as k x dx, and so
x

U=
0

kx dx =

1 2 kx . 2

(12)

Conservation of energy follows from Newtons second law. mv = kx (13)

mv dv = kx dx 1 1 d( mv 2 ) = d( kx2 ) 2 2 1 1 mv 2 + kx2 = constant, 2 2

where we used dx = vdt on the second line, and d(x2 ) = 2x dx and d(v 2 ) = 2v dv on the third line. This means we can write the conservation of energy as E =K +U (14)

and energy constantly sloshes between kinetic and potential energy. We can get an expression for the total energy by starting from the solutions x(t) = A sin(t) v(t) = A cos(t) which means we can write U= 1 1 2 2 kA sin (t) K = mA2 2 cos2 (t) 2 2 (16) (15)

and the total energy is equal to E = = = = K +U 1 2 2 kA sin (t) + 2 1 2 2 kA sin (t) + 2 1 2 kA . 2 1 mA2 2 cos2 (t) 2 1 k mA2 cos2 (t) 2 m (17)

We used sin2 x + cos2 x = 1 in the last line. This means the total energy is proportional to A2 , which is a general feature of oscillators, and sloshes between K and U in the cycle of the oscillator. Why is simple harmonic motion so common? Consider a mass on a spring. We found the potential energy is proportional to x2 , and so the potential energy is a parabola. It is this parabola which gives arise to simple harmonic motion, and is universally important as most oscillations will execute simple harmonic motion when the amplitude is small. Lets prove that. Consider the Taylor series of an arbitrary potential around the point x = 0, which we take as the equilibrium point (a minima). We get U (x) = U (0) + x dU dx 1 + x2 2 x=0 3 d2 U dx2 + ...
x=0

(18)

Now the rst term, U (0) is just a constant and has no signicance for physical observables. So we set it to zero. The second term is zero as the point x = 0 is a minima of the function and so the rst derivitive is zero. So the rst non-zero term (ignoring higher order terms in x, which get small fast for small values of x), the potential, around a minima of an arbitrary potential, has the form U (x) 1 2 x 2 d2 U dx2 .
x=0

(19)

This is just the parabolic behaviour of the potential of a simple harmonic oscillator, and so all potential functions, no matter how complex, have this form close to a minima. The force is then (justifying Hookes law) F = dU = x dx d2 U dx2 .
x=0

(20)

This is why simple harmonic motion is everywhere! For our last topic of section 1.1, consider a pendulum of length l and mass m. For an angular displacement of the mass along the arc of the pendulum, the arc displacement is l. This means the angular velocity along the arc is l and the angular acceleration along the arc is l. The component along the arc of the restoring force, at angle is Frestore = mg sin giving the equation of motion as ml = = mg sin g sin . l (21)

(22)

For small we can expand the sine to get sin() and retain only the rst term to get g = l which looks like our simple harmonic oscillator equation, with solution (t) = 0 cos(t + ) where 2 = g/l and so T = 2 l/g. (25) (24) 3 + ... 3! (23)