Q.A mass of 2 kg is attached to the spring of spring constant $50\ \mathrm{Nm^{-1}}$. The block is pulled to a distance of 5cm from its equilibrium position at $x = 0$ on a horizontal frictionless surface from rest at $t = 0$. Write the expression for its displacement at anytime $t$.

The key idea is that the block executes simple harmonic motion because the spring force is linear ($F = -kx$) and the surface is frictionless. The general solution for displacement is $x(t) = A \sin(\omega t + \phi)$ or $x(t) = A \cos(\omega t + \phi)$.

Step 1: Find angular frequency.

$\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{50}{2}} = \sqrt{25} = 5\ \text{rad/s}$.

Step 2: Identify amplitude and initial condition.

Amplitude $A = 5\ \text{cm} = 0.05\ \text{m}$. …

The problem gives you a mass on a spring on a frictionless surface — a textbook simple harmonic oscillator. When you pull the block to 5 cm and release it from rest, you are giving it maximum displacement with zero initial velocity. That is the classic "cosine" start: the motion begins at the extreme position.

The key idea is that for SHM, the displacement can always be written as $x(t) = A \cos(\omega t + \phi)$ or $x(t) = A \sin(\omega t + \phi)$, depending on the initial conditions. The amplitude $A$ is the maximum displacement from equilibrium, and the phase constant $\phi$ is fixed by where the motion starts.

1. Find the angular frequency. For a spring-mass system, $\omega = \sqrt{\frac{k}{m}}$. Here $k = 50\ \mathrm{N/m}$ and $m = 2\ \mathrm{kg}$, so

2. Identify the amplitude. The block is pulled to 5 cm from equilibrium and released. That distance is the maximum displacement, so

3. Determine the phase constant from initial conditions. At $t = 0$, the block is at $x = +A$ (pulled to the right, say) and released from rest, meaning velocity $v(0) = 0$. Using the cosine form $x(t) = A \cos(\omega t + \phi)$:
   • At $t = 0$, $x(0) = A \cos \phi = A$, so $\cos \phi = 1$, giving $\phi = 0$.
   • Velocity is $v(t) = -A\omega \sin(\omega t + \phi)$. At $t = 0$, $v(0) = -A\omega \sin \phi = 0$, which is automatically satisfied when $\sin \phi = 0$. So $\phi = 0$ works perfectly. …

Step 1: Angular frequency of the spring-mass system: $\omega = \sqrt{k/m} = \sqrt{50/2} = 5\ \text{rad/s}$.

Step 2: Amplitude equals the pull-back distance since the block starts from rest there: $A = 5\ \text{cm} = 0.05\ \text{m}$.

Step 3: Starting from rest at the extreme position ($x(0)=A$, $v(0)=0$) is the standard "pure cosine" initial condition, so $x(t) = A\cos(\omega t)$ with zero phase constant. …