Differential Equations Second Order Help

Question

An undampened spring-mass system with external driving force is modeled with x"+25x = 4 cos 5t. The parameters of this equation are "tuned" so that the frequency of the driving force equals the natural frequency of the undriven system. Suppose that the mass is displaced one positive unit and released from rest.

Find the position of mass as a function of time. What part of the solution guarantees that this solution resonates rather than showing the "beating" character of previous exercises?

Answer 1

We are given the second-order linear nonhomogeneous differential equation modeling an undamped spring-mass system with a resonant driving force:

x′′+25x=4cos⁡(5t)with initial conditions:

x(0)=1,x′(0)=0

Step 1: Identify natural frequency and forcing frequency

The equation:

x′′+25x=0 has natural frequency:

ω0=25=5

The driving force is 4cos⁡(5t), which has frequency:

ω=5

So, forcing frequency = natural frequency, which means the system is in resonance.

Step 2: General solution of the homogeneous equation

The homogeneous equation is:

xh′′+25xh=0

The general solution is:

xh(t)=C1cos⁡(5t)+C2sin⁡(5t)

Step 3: Find a particular solution

Because the forcing function 4cos⁡(5t) has the same frequency as the homogeneous solution, we cannot use Acos⁡(5t)+Bsin⁡(5t)A as a particular solution — it would duplicate terms from the homogeneous solution.

Instead, we use resonant forcing method, multiplying by t:

xp(t)=t(Acos⁡(5t)+Bsin⁡(5t))

Now plug into the equation and solve for AAA and BBB.

Step 4: Compute derivatives

Let’s plug xp=t(Acos⁡(5t)+Bsin⁡(5t)) into the left-hand side:

We find:

xp′′+25xp=4cos⁡(5t)x_p'' + 25

Solving this, you’ll find that:

xp(t)=−(2/5)tsin⁡(5t) (This is a standard result when driving with cos⁡(ωt)\cos(\omega t)cos(ωt) at resonance.)

Step 5: General solution

Now combine the homogeneous and particular solutions:

x(t)=C1cos⁡(5t)+C2sin⁡(5t)−(2/5)tsin⁡(5t)

Apply initial conditions:

1. x(0)=C1=1
2. x′(t)=−5C1sin⁡(5t)+5C2cos⁡(5t)−(2/5)sin⁡(5t)−2tcos⁡(5t)
3. At t=0: x′(0)=5C2=0⇒C2=0

So final solution:

x(t)=cos⁡(5t)−(2/5)tsin⁡(5t)

Step 6: Resonance vs. Beating

The term that guarantees resonance is:

−(2/5)tsin⁡(5t)

This term grows linearly with time, meaning the amplitude of oscillation increases indefinitely — a hallmark of resonance.

In contrast, beating occurs when the driving frequency is close but not equal to the natural frequency, resulting in an amplitude that varies periodically (envelope oscillations), but remains bounded.