Each end of the cart shown above is attached via a spring to the corresponding end of the track. When you displace the cart toward either end of the track and release it, it oscillates about the middle of the track. (The two springs have equal spring constants and free lengths.) As shown in the photograph, the cart is carrying two iron bars. By removing one or both of these, you can change the mass of the cart, and thus the frequency at which the system oscillates. Each bar has a mass of 500 g, and the cart itself has a mass of 500 g, so you can choose to operate the apparatus with the cart having a mass of 500 g, 1 kg or 1.5 kg.
In the apparatus in the photograph above, a cart rides in a pair of grooves on a track. Each end of the cart is connected to the corresponding end of the track via a spring. The two springs are of equal free length and have equal spring constants. When the system is in equilibrium, as shown in the photograph, both springs are extended and apply equal and opposite forces to the cart. The net force on the cart is zero, and it sits at the middle of the track.
When you extend or compress a spring, the force you exert to do so is F = kx, where x is the distance by which you have extended or compressed the spring, relative to its original length, and k is the force constant of the spring. k is a measure of the stiffness of the spring, and its units are N/m. The force the spring exerts in turn is F = -kx, which opposes the force you apply. This is Hooke’s Law.
When you displace the cart from the middle by some distance x, you extend one spring further while compressing the other (or allowing it to relax by the distance x). The force exerted by the springs to oppose the force you apply is the sum of the changes in the force exerted by each spring. The effective force constant of the two springs together is thus the sum of the two individual force constants. Since both springs have the same force constant (about 3.3 N/m), the force constant for the system is twice this (about 6.6 N/m). The analysis of this sytem is thus identical to that of the hanging mass-spring in demonstration 40.12 -- Mass-springs with different spring constants and masses.
We can write F = -kx as
m(d2x/dt2) + kx = 0,
where m is the mass of the cart. This is a second-order differential equation. If we start the system by pulling the cart to one side to a position x0 and then releasing it, then the intitial conditions are that at t = 0, x0 = 0 and dx/dt = 0. If we divide this 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 cart 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 cart, 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 the cart-spring system oscillates is proportional to √k, and inversely proportional to √m. As noted above, k = 6.6 N/m, and for m the three choices are 1.5 kg, 1.0 kg and 0.50 kg. For these values, the calculated frequencies of oscillation are 0.33 Hz, 0.41 Hz and 0.58 Hz, respectively. A rough measurement, obtained by timing five cycles for each mass, gives 0.34 Hz, 0.41 Hz and 0.57 Hz, which are quite close to the calculated frequencies.
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 cart-spring system is the sum of its potential energy and its kinetic energy. The potential energy is the energy stored in the springs. At either turning point, the potential energy is at its maximum. At the equilibrium position, the potential energy is zero, and displacing the cart from there in either direction increases it by (1/2)kx2. 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.