A block of iron having a mass of 100 gram causes a spring to stretch 39.2 cm . If the spring with this mass attached is set into vertical oscillation with amplitude of 15.0 c what is the period of the motion? Neglect the mass of the spring. What is the maximt speed of the block?

To solve this problem, we can use the principles of simple harmonic motion (SHM) and Hooke's law. ### Step 1: Find the spring constant (k) The force exerted by the block of iron due to its weight is given by: \[ F = mg \] where: - \( m = 100 \, \text{g} = 0.1 \, \text{kg} \) (convert grams to kilograms) - \( g = 9.81 \, \text{m/s}^2 \) (acceleration due to gravity) Calculating the force: \[ F = 0.1 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 0.981 \, \text{N} \] According to Hooke's law, the force exerted by a spring is also given by: \[ F = kx \] where: - \( k \) is the spring constant - \( x = 39.2 \, \text{cm} = 0.392 \, \text{m} \) (convert centimeters to meters) Setting the two expressions for force equal to each other: \[ 0.981 \, \text{N} = k \times 0.392 \, \text{m} \] Solving for \( k \): \[ k = \frac{0.981 \, \text{N}}{0.392 \, \text{m}} \approx 2.5 \, \text{N/m} \] ### Step 2: Find the period of the motion (T) The period \( T \) of a mass-spring system is given by the formula: \[ T = 2\pi \sqrt{\frac{m}{k}} \] Substituting the values we have: \[ T = 2\pi \sqrt{\frac{0.1 \, \text{kg}}{2.5 \, \text{N/m}}} \] Calculating: \[ T = 2\pi \sqrt{0.04} = 2\pi \times 0.2 \approx 1.2566 \, \text{s} \] ### Step 3: Find the maximum speed (v_max) The maximum speed in simple harmonic motion can be calculated using the formula: \[ v_{\text{max}} = A \omega \] where: - \( A = 15.0 \, \text{cm} = 0.15 \, \text{m} \) (convert centimeters to meters) - \( \omega = \frac{2\pi}{T} \) First, we need to calculate \( \omega \): \[ \omega = \frac{2\pi}{T} \approx \frac{2\pi}{1.2566} \approx 5.0 \, \text{rad/s} \] Now, substituting \( A \) and \( \omega \) into the equation for maximum speed: \[ v_{\text{max}} = 0.15 \, \text{m} \times 5.0 \, \text{rad/s} = 0.75 \, \text{m/s} \] ### Summary of Results - The period of the motion \( T \approx 1.26 \, \text{s} \) - The maximum speed of the block \( v_{\text{max}} \approx 0.75 \, \text{m/s} \)