Let's solve the problem **step by step**. --- ## **Given:** * A block is hanging in equilibrium from a vertical spring. * When a **second identical block** is added, the **spring sags by 5.00 cm = 0.0500 m**. * We are to find the **oscillation frequencies** for: * (a) **Two-block system**. * (b) **Single-block system**. --- ## ????Step 1: Understand the Physical System This is a vertical spring-mass oscillator. When the mass is hanging at rest, it stretches the spring due to gravity. When it oscillates, it behaves like a **simple harmonic oscillator** with angular frequency: $$ \omega = \sqrt{\frac{k}{m}}, \quad f = \frac{1}{2\pi}\sqrt{\frac{k}{m}} $$ Where: * $k$ = spring constant * $m$ = mass of the system * $f$ = frequency of oscillation * $\omega = 2\pi f$ is the angular frequency --- ## ????Step 2: Analyze the Additional Stretch The second block causes the spring to stretch an **extra 5.00 cm**, so that additional weight equals the spring force: $$ \Delta x = 0.0500\ \text{m} $$ $$ \text{Additional Force} = mg = k \Delta x \Rightarrow k = \frac{mg}{\Delta x} \tag{1} $$ Let $m$ be the mass of **one block**. --- ## ????Step 3: (a) Frequency of the Two-Block System The total mass is now **2m**. Use: $$ f = \frac{1}{2\pi} \sqrt{\frac{k}{2m}} $$ Substitute $k$ from (1): $$ f = \frac{1}{2\pi} \sqrt{\frac{\frac{mg}{\Delta x}}{2m}} = \frac{1}{2\pi} \sqrt{\frac{g}{2 \Delta x}} $$ is the proceed wrong