Let's break down the problem step by step as you requested. Since you mention "Figure 1" but I can't see your plot, I'll use a **typical simple harmonic oscillator (SHO) position vs. time plot** for explanation. If you provide the actual figure, I can be more specific, but this step-by-step will help for any basic SHO plot. ### a. **Write the equation of motion using values from the figure.** #### **Assumptions (typical SHO plot):** - The plot shows a **cosine** or **sine** wave, typical of SHO. - The **amplitude** (maximum displacement) is \(A\). - The **period** (time for one complete cycle) is \(T\). - The **equilibrium position** is at \(x = 0\). - Let's suppose at \(t = 0\), \(x = A\) (maximum), so the motion is described by a **cosine function**. #### **General Equation:** \[ x(t) = A \cos(\omega t + \phi) \] where: - \(A\) = amplitude (meters) - \(\omega\) = angular frequency (radians/s) - \(\phi\) = phase constant #### **Step 1: Find Amplitude \(A\)** From the plot, amplitude is the maximum displacement from equilibrium. **Suppose:** \(A = 0.10~\text{m}\) (example value; use your plot's value) #### **Step 2: Find Period \(T\)** From the plot, measure the time between two adjacent peaks (maximums). **Suppose:** \(T = 4~\text{s}\) #### **Step 3: Find Angular Frequency \(\omega\)** \[ \omega = \frac{2\pi}{T} \] \[ \omega = \frac{2\pi}{4~\text{s}} = \frac{\pi}{2}~\text{rad/s} \] #### **Step 4: Find Phase Constant \(\phi\)** If at \(t = 0\), \(x = A\), then \(\phi = 0\). #### **Step 5: Write the Final Equation** \[ x(t) = 0.10 \cos\left(\frac{\pi}{2} t\right) \] **This is your equation of motion.** --- ### b. **Redraw the position vs. time plot, with labels** #### **Key points to label:** - Mark \(t = 2~\text{s},~4~\text{s},~6~\text{s},~8~\text{s},~10~\text{s}\) on the time axis. - At each, determine: - **i. Direction of motion** (\(+x\) or \(-x\)) - **ii. Sign of velocity** - **iii. Speeding up/slowing down** #### **Let's use the equation above:** \[ x(t) = 0.10 \cos\left(\frac{\pi}{2} t\right) \] #### **Calculate positions:** - \(t=0\): \(x(0) = 0.10 \cos(0) = 0.10\) (max right) - \(t=2\): \(x(2) = 0.10 \cos(\pi) = -0.10\) (max left) - \(t=4\): \(x(4) = 0.10 \cos(2\pi) = 0.10\) (max right) - \(t=6\): \(x(6) = 0.10 \cos(3\pi) = -0.10\) (max left) - \(t=8\): \(x(8) = 0.10 \cos(4\pi) = 0.10\) (max right) - \(t=10\): \(x(10) = 0.10 \cos(5\pi) = -0.10\) (max left) #### **Velocity:** Velocity is the derivative: \[ v(t) = -A\omega \sin(\omega t) \] \[ v(t) = -0.10 \times \frac{\pi}{2} \sin\left(\frac{\pi}{2} t\right) \] - At \(x\) max (\(t=0,4,8\)), \(v=0\) - At \(x\) min (\(t=2,6,10\)), \(v=0\) - In between, velocity alternates in sign. #### **Speeding up/slowing down:** - **Speeding up:** when velocity and acceleration have the same sign - **Slowing down:** when velocity and acceleration have opposite signs Acceleration: \[ a(t) = -A\omega^2 \cos(\omega t) \] --- #### **Summary Table:** | Time (s) | Position | Direction (\(x(t)\) increasing or decreasing) | Velocity Sign | Speeding up/slowing down | |----------|----------|----------------------------------------------|---------------|--------------------------| | 0 | +A | Decreasing (to 0) | 0 (switches -)| Slowing down (stop) | | 1 | 0 | Decreasing (to -A) | - | Speeding up | | 2 | -A | Increasing (to 0) | 0 (switches +)| Slowing down (stop) | | 3 | 0 | Increasing (to +A) | + | Speeding up | | 4 | +A | Decreasing (to 0) | 0 (switches -)| Slowing down (stop) | | 5 | 0 | Decreasing (to -A) | - | Speeding up | | 6 | -A | Increasing (to 0) | 0 (switches +)| Slowing down (stop) | | 7 | 0 | Increasing (to +A) | + | Speeding up | | 8 | +A | Decreasing | 0 (switches -)| Slowing down (stop) | | 9 | 0 | Decreasing | - | Speeding up | | 10 | -A | Increasing | 0 (switches +)| Slowing down (stop) | - **Direction:** If position is moving from +A to -A, direction is \(-x\). If from -A to +A, direction is \(+x\). - **Velocity:** Positive when moving toward \(+x\), negative when toward \(-x\). - **Speeding up:** When moving away from equilibrium (since force and velocity are in same direction). - **Slowing down:** When moving toward turning points (since force and velocity are opposite). --- ### c. **How could the mass be at rest before \(t=0\) and then start oscillating?** **Creative explanations:** - **Scenario 1: A Sudden Push (Impulse)** - The mass was at rest at equilibrium. Suddenly, at \(t=0\), someone gives it a quick push or pull, setting it into oscillation. - **Scenario 2: Release from Displacement** - The mass was held at \(x = +A\) (maximum displacement), at rest. At \(t=0\), it is released, and it starts oscillating. - **Scenario 3: External Disturbance** - An outside force (e.g., a gust of wind, or a tap) acts briefly at \(t=0\), giving the mass some initial energy to start oscillating. - **Scenario 4: Electromagnetic Trigger** - The mass is held still by an electromagnet. At \(t=0\), the electromagnet turns off, and the mass is free to move, starting SHO. **Physical Explanation:** - For the mass to be at rest before \(t=0\) and then oscillate, it must receive **initial conditions** (velocity or displacement) at \(t=0\) to start the motion. This can happen by giving it a sudden displacement (release from rest at max displacement) or a sudden velocity (a tap at equilibrium). --- ## **Final Answers Summary** ### **a. Equation of motion:** \[ \boxed{x(t) = 0.10 \cos\left(\frac{\pi}{2} t\right)} \] *(use the actual amplitude and period from your plot if different)* --- ### **b. Plot labels:** - At \(t = 0, 4, 8~\text{s}\): \(x = +A\) (rightmost), velocity = 0, about to move left (\(-x\)), slowing down near max. - At \(t = 2, 6, 10~\text{s}\): \(x = -A\) (leftmost), velocity = 0, about to move right (\(+x\)), slowing down near min. - In between, motion direction and velocity sign alternate. - **Speed