# Elastic Spring Forces: Mathematical Solution

Example 1:

A block with mass in = 0.76 kg is attached to a spring hanging from the ceiling so that the spring stretched by 12 cm. Find the period of oscillations when the block is stretched 3 cm more.

Solution: We find the spring constant from F = mg and x0 = 0.12 m

k = mg/x0 = (0.76 kg • 9.8 m/s2) / 0.12 m = 62.1 N/m

For x = 0.15 cm = x0 + Δx with Δx = 0.03 m, we have from Newton’s second law:

ma = – kx + mg= -k Δx – k x0 + mg = – k Δx.

We thus find the period:

T = 2π √ (0.76 kg / 62.1 N/m) = 0.7 s.

Elastic Potential Energy, we defined change in potential energy as the negative of the work done on a system. We stretch a spring from x to x + Δx by applying the external force Fext = – Felastic = kx Thus the work done by the elastic work is Welastic = Felastic • Δx = – kx Δx . If we stretch the spring from x = 0 to a finite value x, the average elastic force is Welastic = Felastic,ave * x = (0 – kx) / 2 = – kx/2. so that the work done by the elastic spring forces are Welastic = Felastic,ave • x = (- kx/2) • x = kx2/2. We thus find the elastic potnetial energy:

EPE = kx2/2.

Example 2:

A 2.5 kg block is resting on a flat horizontal table. A horizontal spring with spring constant k = 62 N/m is attached to the block, as shown above. The block is displaced by 8.5 cm to the right and then released from rest. The block then begins to move to the left.

a) Find the time that elapses until the block begins to move to the right.

b) What is the average speed of the block from the moment it is released until it begins to move to the right?

c) What is the total energy of the system [i.e., the spring plus the mass]?

d) Assume that the block is released from rest at time t = 0. Find the time when the elastic energy of the spring is three times the kinetic energy of the block.

Solution: The period of the oscillation is

T = 2π √(m/k) = = 2π √( 2.5 kg /62 N/m) = 1.26s

The time when the block is moving to the left is half the period: t* = T/2 = 0.63 s. The total displacement is Δx = 2A = 2.0.085 m = 0.17 m. Then

Vave = Δx/ t* = 0.17m / 0.63s = 0.27 m/s

The kinetic energy is zero when the block is released. The total mechanical energy follows

Etot = EPE = ½ kA2 = ½ 62 N/m (0.085 m)2 = 0.22 J.

When EPE = 3KE, the total mechanical energy follows Emech = KE + EPE = 4KE, so that,

KE = ½ mv² = Etot/4 = 1/8 kA= 1/8 mω²A²,

where we inserted k = mω², we find

v = ± ½ ωA.

Since v(t) = − Aω sin(ωt), we get

− Aω sin(ωt) = − Aω/2 so that sin(ωt) = ½. we find