91Ó°ÊÓ

Particle acceleration describes how quickly the velocity of a particle changes. In this problem, the acceleration is proportional to the distance from a fixed point, C. This is known as simple harmonic motion.

Simple harmonic motion can be described by the equation:
\[ a = -\beta x \]
The negative sign indicates that the acceleration is always directed towards point C (restoring force).

For this specific problem, the constants are given as \(a = -\beta x\). We substitute \(\beta\) with \(\beta = \lambda^2\), giving so we have:
\[ a = -\lambda^2 x \]
On top of that we know that the particle starts from rest at distance \(d\) from point \(C\), overall giving:
\[ a = -\lambda^2 x \] (initial condition).

In physics problems involving motion, differential equations play a critical role. A differential equation relates a function with its derivatives. In this exercise, we aim to solve a second-order differential equation because acceleration, which is the second derivative of position, is involved.

Starting with the given acceleration equation:
\[ a = -\lambda^2 x \]
We use the fact that acceleration (\(a\)) is the second derivative of position \(x\) with respect to time (\(t\)):
\[ a = \frac{d^2x}{dt^2} = -\lambda^2 x \]

This sort of differential equation is characteristic of simple harmonic motion. We solve this by assuming a solution of the form:
\[ x(t) = A \cos(\lambda t) + B \sin(\lambda t) \]
where \(A\) and \(B\) are constants determined by initial conditions.

Initial conditions help in determining the specific solution to our differential equation. For this exercise, the particle starts from rest at a distance \(d\) from point \(C\). These initial conditions (at \(t = 0\)) are:
* \(x(0) = d\) (particle is at point A)
* \(v(0) = 0\) (particle starts from rest)

With the initial condition \(x(0) = d\), we can solve for constants in our assumed solutions and find the specific trajectory of the particle. Starting with our general solution:
\[ x(t) = A \cos(\lambda t) + B \sin(\lambda t) \]
Applying \(x = d\) at \(t = 0\) gives:
\[ d = A \cos(0) + B \sin(0) \]
Since \(\cos(0)=1\) and \(\sin(0)=0\), we get \(A = d\). So, our equation becomes:
\[ x(t) = d \cos(\lambda t) \]
Using the initial condition \(v(0) = 0\):
\[ v(t) = \-d\ \lambda\ \sin(\lambda t) \]
Thus, our complete solutions for position and velocity as functions of time is:
\[ x(t) = d \cos(\lambda t) \]
\[ v(t) = -d \ \lambda \ \sin(\lambda t) \]