Let's work through each question step by step, as requested. --- ### (a) What is the amplitude \( A \) of this system’s simple harmonic motion? **Step 1:** Identify the equation of motion given: \( x = (.384\, \text{m}) \cos \left[ (4.72\, \text{s}^{-1})t \right] \) **Step 2:** The amplitude is the coefficient in front of the cosine function. **Explanation:** In the equation \( x = A \cos(\omega t) \), the value \( A \) is the amplitude. **Final Answer:** **Amplitude, \( A = .384\, \text{m} \)** --- ### (b) What is the spring constant of this spring? **Step 1:** The angular frequency \( \omega = 4.72\, \text{s}^{-1} \) is given. For a mass-spring system, \( \omega = \sqrt{\dfrac{k}{m}} \). **Step 2:** Rearrange to solve for \( k \): \( k = m\omega^2 = (.650\, \text{kg}) \times (4.72\, \text{s}^{-1})^2 \) \( k = .650 \times 22.2784 = 14.48196\, \text{N/m} \) **Explanation:** The spring constant \( k \) relates to the mass and angular frequency by the above formula. **Final Answer:** **Spring constant, \( k = 14.5\, \text{N/m} \) (rounded to 3 sig figs)** --- ### (c) What is the maximum speed of the block? **Step 1:** Maximum speed \( v_{\text{max}} = \omega A \). **Step 2:** \( v_{\text{max}} = 4.72\, \text{s}^{-1} \times .384\, \text{m} = 1.81248\, \text{m/s} \) **Explanation:** Maximum speed in SHM is the product of angular frequency and amplitude. **Final Answer:** **Maximum speed, \( v_{\text{max}} = 1.81\, \text{m/s} \) (rounded to 3 sig figs)** --- ### (d) Write the equation for the velocity of the block at time \( t \) only. **Step 1:** The general velocity equation for SHM: \( v(t) = -A\omega \sin(\omega t) \) **Step 2:** Substitute the values: \( v(t) = -.384\, \text{m} \times 4.72\, \text{s}^{-1} \sin(4.72t) \) \( v(t) = -1.81 \sin(4.72t) \) m/s **Explanation:** Velocity is the derivative of the position function with respect to time. **Final Answer:** **\( v(t) = -1.81 \sin(4.72t) \) m/s** --- ### (e) Find the period of this system’s simple harmonic motion. **Step 1:** The period \( T = \dfrac{2\pi}{\omega} \). **Step 2:** \( T = \dfrac{2\pi}{4.72} = 1.332 \, \text{s} \) **Explanation:** The period is the reciprocal of the frequency, which is related to angular frequency. **Final Answer:** **Period, \( T = 1.33\, \text{s} \) (rounded to 3 sig figs)** --- Let me know if you want further clarification on any step!