Let's break down and solve the problem step by step: --- ## **Given Data** - Spring stiffness, \( k = 4 \) N/m - Mass, \( m = 1 \) kg - Rod length, \( l = 1 \) m - Spring attached at distance, \( a = .5 \) m from pivot - Initial rotation angle, \( \theta_ = .2 \) radians - Time, \( t = 5 \) s - Moment of inertia, \( J = ml^2 \) - Position equation: \( \theta(t) = A \sin(\omega_n t + \phi) \) - **Find:** The vertical distance (in m) between the mass and point O after 5 seconds. --- ## **Step 1: Find the Natural Frequency \(\omega_n\)** The spring creates a **restoring torque** about the pivot: \[ \tau = -k a^2 \theta \] The equation of motion: \[ J \ddot{\theta} + k a^2 \theta = \] where \( J = ml^2 \). So, \[ ml^2 \ddot{\theta} + k a^2 \theta = \] \[ \ddot{\theta} + \frac{k a^2}{ml^2} \theta = \] \[ \omega_n = \sqrt{\frac{k a^2}{ml^2}} \] Plug in the values: - \( k = 4 \) N/m - \( a = .5 \) m - \( m = 1 \) kg - \( l = 1 \) m \[ \omega_n = \sqrt{\frac{4 \cdot (.5)^2}{1 \cdot (1)^2}} = \sqrt{\frac{4 \cdot .25}{1}} = \sqrt{1} = 1 \text{ rad/s} \] --- ## **Step 2: Set Up the Position Equation** Given initial rotation \(\theta_ = .2\) radians and assuming initial angular velocity is zero: \[ \theta() = A \sin(\phi) = .2 \] \[ \dot{\theta}() = A \omega_n \cos(\phi) = \] So, \(\cos(\phi) = \Rightarrow \phi = \frac{\pi}{2}\). \[ \theta() = A \sin\left(\frac{\pi}{2}\right) = A \cdot 1 = .2 \implies A = .2 \] So, \[ \theta(t) = .2 \sin(1 \cdot t + \frac{\pi}{2}) \] \[ \sin(x + \frac{\pi}{2}) = \cos(x) \] So, \[ \theta(t) = .2 \cos(t) \] --- ## **Step 3: Find \(\theta\) at \(t = 5\) s** \[ \theta(5) = .2 \cos(5) \] Calculate \(\cos(5)\), where \(5\) is in radians: \[ \cos(5) \approx .28366 \] \[ \theta(5) = .2 \times .28366 = .05673 \text{ radians} \] --- ## **Step 4: Find the Vertical Distance** At any time, the mass is at the end of the rod, length \(l = 1\) m. The **vertical distance** from O to the mass is: \[ \text{vertical distance} = l (1 - \cos(\theta)) \] Plug in values: \[ l = 1 \text{ m},\ \theta(5) = .05673 \] \[ \cos(.05673) \approx .99839 \] \[ \text{vertical distance} = 1 \cdot (1 - .99839) = .00161 \text{ m} \] --- ## **Final Answer** \[ \boxed{.00161\ \text{m}} \] **The vertical distance between the mass and the point O after 5 seconds is .00161 m.**