The given equation is $x = A \cos(p t - \alpha)$, which describes a harmonic oscillator, where:

• $A$ is the amplitude,
• $p$ is the angular frequency, and
• $\alpha$ is the phase shift.

To form a differential equation from this, we differentiate with respect to time ($t$) and eliminate the constants to obtain a relationship involving $x$ and its derivatives.

Step 1: First Derivative

The first derivative of $x$ with respect to time gives the velocity: $\frac{dx}{dt} = \frac{d}{dt} \left( A \cos(p t - \alpha) \right) = -A p \sin(p t - \alpha)$

Step 2: Second Derivative

The second derivative of $x$ with respect to time gives the acceleration: $\frac{d^2x}{dt^2} = \frac{d}{dt} \left( -A p \sin(p t - \alpha) \right) = -A p^2 \cos(p t - \alpha)$

Since $x = A \cos(p t - \alpha)$, we can substitute $x$ back into this second derivative equation:

$\frac{d^2x}{dt^2} = -p^2 x$

Step 3: Final Differential Equation

The resulting differential equation is:

$\frac{d^2x}{dt^2} + p^2 x = 0$

This is a second-order linear homogeneous differential equation, characteristic of simple harmonic motion.

Would you like more details or further explanation?

Here are 5 related questions you could explore:

1. How does the solution of this differential equation describe simple harmonic motion?
2. What is the physical meaning of the angular frequency $p$ in this context?
3. How would the equation change if there was damping involved in the system?
4. What initial conditions are necessary to fully determine the constants $A$ and $\alpha$?
5. How would you solve this equation for different boundary conditions?

Tip: The general form of a simple harmonic motion equation is always of the type $\frac{d^2x}{dt^2} + \omega^2 x = 0$, where $\omega$ is the angular frequency.