mass spring example of oscillation

Oscillations in Physics: A Detailed Worked Mass Spring Example

This article develops one problem in enough detail to connect most of the basic quantities introduced in the PhysicsLibrary entry on oscillations. The goal is not merely to obtain x(t), but to interpret the amplitude, phase, period, frequency, velocity, acceleration, turning points, energy exchange, and phase-space trajectory as parts of one physical oscillation.

The example is deliberately given nonzero initial displacement and nonzero initial velocity. This prevents the phase constant from being hidden by a specially chosen initial condition and makes the general sinusoidal solution physically meaningful.

Problem

A block of mass

m = 0.50 kg

slides without friction on a horizontal surface and is attached to an ideal spring of spring constant

k = 18N ∕m.

Let x = 0 denote the equilibrium position and let positive x point to the right. At t = 0 the block is at

x0 = +0.060 m

and is moving to the right with velocity

v0 = +0.48 m ∕s.

Determine and interpret the complete oscillation. In particular:

derive the equation of motion from Newton’s second law;
find the natural angular frequency, period, and frequency;
solve for x(t) using the initial conditions;
express the result in amplitude-phase form and determine A and ϕ;
obtain v(t) and a(t);
find the maximum speed and maximum acceleration;
find the first positive turning point, the first equilibrium crossing, and the first negative turning point after t = 0;
calculate the total mechanical energy and verify it using the initial conditions;
show how kinetic and potential energy exchange during the oscillation;
derive the phase-space ellipse in the (x,v) plane.

Figure 1. Mass spring system at t = 0. The block begins to the right of equilibrium and is initially moving farther to the right, while the spring restoring force points back toward equilibrium.

Solution

1. Derive the equation of motion

The spring force obeys Hooke’s law,

Fs = − kx.

The minus sign is important. If x > 0, the spring force points toward negative x; if x < 0, the spring force points toward positive x. The force is therefore restoring.

Newton’s second law gives

m ¨x = − kx.

Move all terms to the left:

m ¨x + kx = 0.

Dividing by m,

x¨+ k-x = 0. m

Define

ω20 = k-. m

Then the equation becomes

2 x¨+ ω 0x = 0.

This is the differential equation of an undamped simple harmonic oscillator.

2. Natural angular frequency, period, and frequency

For the given system,

∘ --- ∘ ----- k 18 √ --- ω0 = -- = ---- = 36 = 6.00 rad∕s. m 0.50

The period is

2π 2π π T = --- = --- = --≃ 1.047s. ω0 6 3

The frequency is

1 ω0 f = -- = --- ≃ 0.955Hz. T 2π

Thus the block completes slightly less than one full cycle each second.

3. Solve the differential equation with the initial conditions

A convenient general solution is

x(t) = C cos(ω0t ) + D sin(ω0t).

With ω₀ = 6, this becomes

x(t) = C cos(6t) + D sin(6t).

At t = 0,

x(0) = C.

Since x(0) = 0.060 m,

C = 0.060m.

Differentiate the general solution:

v (t) = ˙x(t) = − 6C sin (6t) + 6D cos(6t).

At t = 0,

v(0) = 6D.

Therefore

v0 0.48 D = -- = ---- = 0.080 m. 6 6

The motion is therefore

x (t) = 0.060 cos(6t) + 0.080 sin(6t) m.

This form is already a complete solution. The amplitude-phase form, however, makes the geometry of the oscillation easier to read.

4. Convert to amplitude-phase form

Write the same motion as

x(t) = A cos(ω t + ϕ). 0

Using

cos(ω0t + ϕ ) = cos(ω0t)cosϕ − sin(ω0t)sin ϕ,

we identify

C = A cos ϕ,

and

D = − A sin ϕ.

Squaring and adding gives

2 2 2 A = C + D .

Thus

∘ -------2----------2 A = (0.060) + (0.080) = 0.100 m.

Now

C cosϕ = --= 0.60, A

while

sinϕ = − D-= − 0.80. A

Therefore the correct quadrant gives

ϕ = − 0.9273 rad = − 53.13∘.

The complete amplitude-phase solution is

x (t) = 0.100 cos(6t − 0.9273) m.

The negative phase tells us that the oscillator is already partway through its cycle at t = 0. Because the initial velocity is positive, the block has not yet reached its positive turning point.

5. Velocity and acceleration

Differentiate the displacement:

v(t) = ˙x(t) = − 0.600 sin (6t − 0.9273) m∕s.

Differentiate again:

2 a(t) = ¨x (t) = − 3.600cos(6t − 0.9273) m ∕s .

Because

x (t) = 0.100 cos(6t − 0.9273),

we can also write

2 a(t) = − ω0x (t) = − 36x (t).

This is the defining kinematic property of simple harmonic motion: acceleration is proportional to displacement and always directed toward equilibrium.

Figure 2. Normalized displacement, velocity, and acceleration over one period. Velocity is shifted by one quarter cycle relative to displacement, while acceleration is exactly opposite in phase to displacement.

6. Maximum speed and maximum acceleration

The velocity has amplitude

v = A ω . max 0

Therefore

v = (0.100)(6.00) = 0.600m ∕s. max

The acceleration has amplitude

2 amax = A ω0.

Thus

2 amax = (0.100)(36) = 3.60 m ∕s.

The maximum speed occurs at equilibrium, where all the mechanical energy is kinetic. The maximum acceleration occurs at the turning points, where |x| = A.

7. Important times during the first cycle

Let the phase be

ψ (t) = 6t − 0.9273.

At t = 0,

ψ(0) = − 0.9273.

Since the block is moving to the right, it first reaches the positive turning point when

ψ = 0.

Therefore

6t − 0.9273 = 0,

t+A = 0.1545 s.

At this instant,

x = +A = +0.100 m, v = 0.

The first equilibrium crossing occurs when

π- ψ = 2.

Thus

π -2 +-0.9273 tx=0 = 6 ≃ 0.4163 s.

At this crossing the block is moving toward negative x, so

v = − vmax = − 0.600 m∕s.

The first negative turning point occurs when

ψ = π.

Therefore

π-+-0.9273- t− A = 6 ≃ 0.6781 s.

At this instant,

x = − 0.100 m, v = 0.

These times show the physical cycle directly: the block first moves farther right, stops, reverses, passes through equilibrium at maximum speed, and then reaches the left turning point.

8. Total mechanical energy

For an ideal horizontal mass spring oscillator,

E = K + U,

where

K = 1mv2 2

and

1- 2 U = 2 kx .

At a turning point v = 0 and |x| = A, so

E = 1-kA2. 2

Using k = 18 N∕m and A = 0.100 m,

E = 1(18)(0.100)2 = 0.0900 J. 2

We can verify this independently from the initial conditions. Initially,

1- 2 1- 2 U0 = 2 kx0 = 2 (18 )(0.060 ) = 0.0324 J,

and

K = 1mv2 = 1-(0.50 )(0.48)2 = 0.0576 J. 0 2 0 2

Thus

E0 = K0 + U0 = 0.0576 + 0.0324 = 0.0900J.

The two calculations agree, as they must for an ideal conservative oscillator.

9. Energy exchange through the cycle

Substituting

x = A cos ψ

into the spring potential gives

1 2 2 U = -kA cos ψ. 2

Because

v = − Aω sinψ, 0

the kinetic energy is

K = 1mA2 ω2sin2 ψ. 2 0

But

m ω20 = k,

K = 1-kA2 sin2 ψ. 2

Adding,

1- 2( 2 2 ) E = 2kA sin ψ + cos ψ .

Therefore

E = 1-kA2, 2

which is constant.

At a turning point, U = E and K = 0. At equilibrium, K = E and U = 0. Between those locations, the energy is shared between the two forms.

Figure 3. Kinetic and spring potential energy exchange twice per oscillation cycle, while their sum remains constant at 0.090 J.

10. Phase-space ellipse

The displacement and velocity are

x = A cosψ,

v = − Aω0 sinψ.

Divide the first equation by A and the second by Aω₀:

-x A = cos ψ,

v ---- = − sin ψ. A ω0

square and add:

2 2 x--+ -v---= 1. A2 A2ω20

For this oscillator,

A = 0.100 m, A ω0 = 0.600m ∕s,

so the phase-space trajectory is

x2 v2 (0.100-)2 + (0.600-)2 = 1.

The oscillator therefore traces a closed ellipse in the (x,v) plane. A closed orbit is another way of representing the periodic return of an ideal conservative oscillator to the same dynamical state.

Figure 4. Phase-space trajectory of the oscillator. The initial state (x₀,v₀) = (0.060,0.480) lies on the same constant-energy ellipse traced throughout the motion.

Physical interpretation of the complete motion

This one example contains most of the basic ideas used to describe an oscillation:

The equilibrium is at x = 0.
The restoring force F = −kx always points toward equilibrium.
The amplitude A = 0.100 m sets the turning points.
The angular frequency ω₀ = 6.00 rad∕s is determined by k∕m, not by the initial displacement or velocity.
The phase constant ϕ = −0.9273 rad records where in the cycle the chosen clock starts.
The period T = 1.047 s is the time required to return to the same position and velocity.
Velocity is largest at equilibrium and zero at the turning points.
Acceleration is zero at equilibrium and largest in magnitude at the turning points.
Energy continuously transfers between kinetic and spring potential energy while the total remains constant.
The closed phase-space ellipse represents the same conserved-energy motion geometrically.

The initial conditions change A and ϕ, but for an ideal linear mass spring oscillator they do not change ω₀. That separation between the properties of the system and the choice of initial state is one of the most useful features of linear oscillation theory.