The two springs at left have the same spring constant; the one at right is stiffer. The two masses on the springs at left and right are equal (500 g); the mass on the middle spring is less massive (200 g). When you set the three mass-springs oscillating, you can show that of the two systems whose springs have the same spring constant, the one with the larger mass exhibits the lower oscillating frequency, and that of the two systems whose masses are equal, the one whose spring has the greater spring constant has the higher frequency of oscillation.
When you extend (or compress) a spring, the force that the spring exerts in response is F = -kx, where x is the distance by which you compress or extend the spring, and k is the spring constant, which is a measure of the stiffness of the spring. This is Hooke’s Law. The net force on each mass hanging in the apparatus shown above is thus F = mg - kx. When the mass-spring is at its equilibrium position, the weight of the mass stretches the spring from its free length by a distance s, and F = mg - ks = 0. If we now displace the mass from the equilibrium position by some distance x, we have F = mg - ks - kx. Since mg = ks, we have F = -kx, or
m(d2x/dt2) + kx = 0.
This is a second-order differential equation. If we start one of the systems by pulling down the mass to a position x0 and then releasing it, then the initial conditions are that at t = 0, x = x0 and dx/dt = 0. If we divide our differential equation by m and substitute ω = √(k/m), we have d2x/dt2 + ω2x = 0. This has the characteristic equation r2 + ω2 = 0, whose roots are r = ±iω. This leads to the general solution
x = C1 cos ωt + C2 sin ωt.
With our initial conditions, C1 = x0, and C2 = 0, and
x = x0 cos ωt.
This equation describes simple harmonic motion whose amplitude is x0, whose frequency in radians per second is ω, which equals √(k/m), and whose period is T = 2π/ω. In cycles per second, the frequency is ω/(2π).
If we set C1 = A cos φ and C2 = A sin φ (where A = √(C12 + C22) and tan φ = C2/C1), we can use the trigonometric identity
A cos(ωt - φ) = A cos φ cos ωt + A sin φ sin ωt
to obtain
x = A cos (ωt - φ)
A is the amplitude of the oscillation, which corresponds to the maximum displacement of the mass from its equilibrium position to either turning point. ω (= √(k/m)) is the vibrational frequency in radians per second, and φ is a phase factor, which depends on the initial speed and position of the mass, as does A. (Whether the function one obtains is a sine or a cosine function depends on the choice of initial conditions, that is, whether x0, the position at t = 0, is the equilibrium position or a turning point.)
From this, we see that the frequency at which a mass-spring system oscillates is proportional to √k, and inversely proportional to √m.
We also find an interesting relationship among x, v and a, the position, velocity and acceleration of the mass. If we take x = A cos (ωt - φ), then v = dx/dt = -ωA sin (ωt - φ), and a = d2x/dt2 = -ω2A cos (ωt - φ). These equations show us that at the extrema (maximum and minimum x), the velocity is zero, and the acceleration is at its maximum, and that at the equilibrium position (x = 0), the acceleration is zero and the velocity is at its maximum.
The energy of the mass-spring system is the sum of its potential energy and its kinetic energy. The potential energy is the combination of gravitational potential energy and the energy stored in the spring. At the top turning point, the gravitational potential energy is at its maximum, and the energy in the stretched spring is at its minimum. At the bottom turning point, the gravitational potential energy is at its minimum, and the energy in the stretched spring is at its maximum. At the equilibrium position, their sum is at its minimum, and displacing the mass from there in either direction increases the potential energy by (1/2)kx2. We can thus take the equilibrium position of the system as the zero reference for the potential energy. The kinetic energy is (1/2)mv2. The total energy (the sum of potential energy and kinetic energy) must remain constant, except for frictional losses. The equations above, for x and v, show that at the turning points, the potential energy is at its maximum, and the kinetic energy equals zero. At the equilibrium position, the potential energy equals zero, and the kinetic energy is at its maximum.
References:
1) Thomas, George B., Jr. and Finney, Ross L. Calculus and Analytic Geometry (Reading, Massachusetts: Addison-Wesley Publishing Company, 1992), pp. 1080-1081.
2) Resnick, Robert and Halliday, David. Physics, Part One, Third Edition (New York: John Wiley and Sons, 1977), pp. 303-308.
3) Sears, Francis Weston and Zemansky, Mark W. College Physics, Third Edition (Reading, Massachusetts: Addison-Wesley Publishing Company, 1960) p. 218-226.
4) http://hyperphysics.phy-astr.gsu.edu/hbase/shm2.html#c2 and links to related topics.