91Ó°ÊÓ

Start with Newton's second law, which states that the sum of the forces acting on an object is equal to the mass of the object times its acceleration. For a block attached to a spring, the only forces that are acting on it are the elastic (spring) force and the external force. The spring force is represented as \( -kx \), where \( k \) is the spring constant and \( x \) is the displacement from the equilibrium position. It's negative because it always acts opposite to the direction of displacement. The external force is given as \( F_0 \cos \omega t \), where \( F_0 \) is the amplitude, \( \omega \) is the angular frequency and \( t \) is time. Set the sum of these forces equal to \( m \ddot{x} \), where \( m \) is the mass of the block and \( \ddot{x} \) is the second derivative of displacement with respect to time (the acceleration). The resulting equation is \( m \ddot{x} = -kx + F_0 \cos \omega t \).

Simplify the equation by dividing each term by \( m \), giving us \( \ddot{x} = -\frac{k}{m}x + \frac{F_0}{m} \cos \omega t \). This is the derived differential equation of motion.

The general solution to this differential equation involves ordinary differential equations and is beyond the scope of the current exercise to derive. However, for completeness, it is stated as \( x(t) = A \cos (\omega t - \delta) + \frac{F_0}{\sqrt{m^2(\omega_0^2 - \omega^2)^2 + (b\omega)^2}} \cos (\omega t - \theta) \), where \( A \) and \( \delta \) are the amplitude and phase angle of the natural response (which depends on initial conditions), \( \theta \) is the phase angle of the steady-state response, \( \omega_0 \) is the natural frequency of the system, and \( b \) is the damping coefficient.