Question

Consider a spring-mass system with mass \( m \) and spring stiffness \( k \) as shown in the illustration. At time \( t = 0 \), the mass is displaced by \( P \) units and the velocity of the mass is zero. The displacement of the mass, \( x(t) \), is measured from the equilibrium position. Which one of the following functions represents \( x(t) \)? 

• A. \( P \cos \left( \sqrt{\frac{k}{m}} t \right) + P \sin \left( \sqrt{\frac{k}{m}} t \right) \)
• B. \( P \sin \left( \sqrt{\frac{k}{m}} t \right) \)
• C. \( \frac{P}{2} \cos \left( \sqrt{\frac{k}{m}} t \right) + \frac{P}{2} \sin \left( \sqrt{\frac{k}{m}} t \right) \)
• D. \( P \cos \left( \sqrt{\frac{k}{m}} t \right) \)

Hint:

For simple harmonic motion systems, the displacement function is generally of the form \( x(t) = A \cos(\omega t) + B \sin(\omega t) \), where \( \omega = \sqrt{\frac{k}{m}} \) and the constants \( A \) and \( B \) are determined by the initial displacement and velocity.

Answer 1

The Correct Option is D

Step 1: For a spring-mass system with mass \( m \) and spring stiffness \( k \), the equation of motion is given by: \[ m \frac{d^2x(t)}{dt^2} + kx(t) = 0 \] This is a second-order linear differential equation, and its solution has the general form: \[ x(t) = A \cos \left( \sqrt{\frac{k}{m}} t \right) + B \sin \left( \sqrt{\frac{k}{m}} t \right) \] where \( A \) and \( B \) are constants to be determined by initial conditions. 
Step 2: Given that at \( t = 0 \), the displacement is \( P \) and the velocity is zero, we apply these initial conditions: \[ x(0) = P {and} \frac{dx}{dt} (0) = 0 \] From the first condition, we get: \[ A = P {and} B = 0 \] Thus, the solution simplifies to: \[ x(t) = P \cos \left( \sqrt{\frac{k}{m}} t \right) \] This matches with option (D).