Question

1. A mechanical system with one degree of freedom oscillates about a stable equilibrium state. Its displacement from the equilibrium, x(t), satisfies the simple harmonics oscillator equation:

d2x + ω2 x = 0.dt2

(a) What is the characteristic period of the oscillation?
(b) Write a solution to the above equation for which x(0) = x0 and ẋ(0) = v0. (c) Demonstrate that E = ẋ 2 + ω 2 x 2 does not vary in time. What is the physical significance of E?

2. An damped mechanical system with one degree of freedom oscillates about a stable equilibrium state. Its displacement from the equilibrium, x(t), satisfies the damped harmonic oscillator equation:

ẍ+νẋ+ω2 x=0, where ν > 0.

(a) Demonstrate that

x = A sin(ω1 t)eγt

is a particular solution of the damped harmonic oscillator equation, and determine the values of ω1 and γ (assuming that ν < 2 ω).

(b) Demonstrate that

dE ≤ 0,dt

where E is defined in Q1(c). What is the physical significance of this equation?

3. Consider a mass-spring system consisting of two identical masses, of mass m, which slid over a frictionless horizontal surface. In order, from the left to the right, the system consists of a spring of spring constant k′ whose left end is attached to an immovable wall and whose right end is attached to the first mass, a spring of spring constant k whose left end is attached to the first mass and whose right end is attached to the second mass, and a spring of spring constant k′ whose left end is attached to the second mass and whose right end is attached to an immovable wall. Let ω0 = √k/m and α = k′/k.

(a) Demonstrate that the equations of motion of the system can be written:

ẍ1 = −(1+α)ω02 x1 +ω02 x2,

ẍ2 = ω02 x1 −(1+α)ω02 x2,

(b) Demonstrate that the normal frequencies of the system are ω = α1/2 ω0 and ω = (2 +α)1/2 ω0.

(c) Demonstrate that the low frequency normal mode is such that the two masses oscillate in phase with the same amplitude, and that the high frequency mode is such that the two masses oscillate in anti-phase with the same amplitude.

Answer 1

1. As per the question, Simple Harmonics Oscillator satisfies following equation:

d²x + ω² x = 0.dt²

or, d²x / dt² = - ω² x

We can find a mathematical solution to above equation. We need a function whose second derivative is the same as the original function with a negative sign and multiplied by ω². We can easily figure out that the trigonometric functions sine and cosine has this characteristics, so we can write either of following equations:

x = A sin(ω t + θ) or x = A cos(ω t + θ)

For further discussion we chose equation x = A sin(ω t + θ)

Part a)

To find the period of oscillation, we can observe that

x = A sin(ω t + θ) will repeat itself when ω t is multiple of 2π as sinθ = sin(2π *n +θ) where n is integer

So, if the Time period of oscillation is T,

then ω T = 2π

so, T = 2π/ ω

Part b)

We will use our equation

x = A sin(ω t + θ)

so, v = dx/dt = ω A cos (ω t + θ)

As per question, for t= 0, x = x₀ and v = v₀

If we put t = 0 in our equation for x and v above , we will find

x₀ = A sin θ

A = x₀ / sin θ

similarly,

v₀ =  ω A cos θ

v₀ =  ω (x_o / sinθ ) cos θ

v₀ =  ω x_o cot θ

cot θ = v₀ / ω x_o

tan θ = ω x_o / v₀

θ = tan^-1 (ω x_o / v₀);

Now, we can rewrite our equations as below:

x = (x₀ / sinθ) sin(ω t + θ) ; where θ = tan^-1 (ω x_o / v₀)

Part c)

Energy of this oscillation can be given as:

Kinetic energy = (1/2) mv²

Potential energy = (1/2) kx²

In case of this oscillation, force ; F = ma = -kx ; a is acceleration ; m is mass of object

and ω² = k/m

We can revisit our equations:

x = A sin(ω t + θ)

v = ω A cos (ω t + θ)

Maximum value of x = A and maximum value of v = ω A

Total Energy is

E = (1/2) mv² + (1/2) kx²

E = (1/2) m ω² A² cos²(ω t + θ) + (1/2) k A² sin²(ω t + θ)

E = (1/2) m (k/m) A² cos²(ω t + θ) + (1/2) k A² sin²(ω t + θ) ; as ω² = k/m

E = (1/2) k A² (cos²(ω t + θ) + sin²(ω t + θ))

E = (1/2) kA²

So, it proves that Energy remains constant throughout.

When v becomes zero, there is no kinetic energy; but potential energy increases with maximum value of x at that point and when x = 0, there is no potential energy and it is in Kinetic form with maximum speed