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

Hint:

Always start with the general SHM solution \(x(t) = A\cos(\omega t) + B\sin(\omega t)\). Use initial displacement → gives \(A\), and initial velocity → gives \(B\).

Answer 1

The Correct Option is D

Step 1: Equation of motion.
The governing differential equation for spring–mass system: \[ m \frac{d^2x}{dt^2} + kx = 0 \] Divide by \(m\): \[ \frac{d^2x}{dt^2} + \frac{k}{m} x = 0 \]

Step 2: General solution.
This is a simple harmonic oscillator. General solution: \[ x(t) = A \cos(\omega t) + B \sin(\omega t) \] where \[ \omega = \sqrt{\tfrac{k}{m}} \]

Step 3: Apply initial condition (1): displacement at \(t=0\).
At \(t=0\): \[ x(0) = A \cos(0) + B \sin(0) = A \] But given \(x(0) = P\). \[ A = P \]

Step 4: Apply initial condition (2): velocity at \(t=0\).
Velocity: \[ v(t) = \frac{dx}{dt} = -A \omega \sin(\omega t) + B \omega \cos(\omega t) \] At \(t=0\): \[ v(0) = B \omega \] But given \(v(0)=0 \Rightarrow B=0\).



Step 5: Final expression.
Thus, \[ x(t) = P \cos \left(\sqrt{\tfrac{k}{m}} t \right) \] Final Answer:
\[ \boxed{P \cos \left(\sqrt{\tfrac{k}{m}} t \right)} \]