1. **State the problem:** A 13.6-kg block is attached to three springs arranged vertically: the top spring with stiffness $k_1=3500$ N/m, and two bottom springs with stiffness $k_2=2100$ N/m and $k_3=2800$ N/m. The block is displaced downward by 44 mm (0.044 m) from equilibrium and released. 2. **Find the effective spring constant:** The two bottom springs are in parallel, so combined stiffness $k_{bottom} = k_2 + k_3 = 2100 + 2800 = 4900$ N/m. 3. The top spring is in series with the bottom combination. The equivalent spring constant $k_{eq}$ satisfies $$ \frac{1}{k_{eq}} = \frac{1}{k_1} + \frac{1}{k_{bottom}} = \frac{1}{3500} + \frac{1}{4900} $$ Calculate: $$ \frac{1}{k_{eq}} = \frac{1}{3500} + \frac{1}{4900} = \frac{4900 + 3500}{3500 \times 4900} = \frac{8400}{17150000} $$ Thus, $$ k_{eq} = \frac{17150000}{8400} \approx 2041.67 \text{ N/m} $$ 4. **Calculate the period $T$ of oscillation:** $$ T = 2\pi \sqrt{\frac{m}{k_{eq}}} = 2\pi \sqrt{\frac{13.6}{2041.67}} $$ Calculate under the square root: $$ \sqrt{\frac{13.6}{2041.67}} \approx \sqrt{0.00666} \approx 0.0816 $$ Then, $$ T \approx 2\pi \times 0.0816 \approx 0.513 \text{ seconds} $$ 5. **Calculate the frequency $f$:** $$ f = \frac{1}{T} \approx \frac{1}{0.513} \approx 1.95 \text{ Hz} $$ 6. **Calculate the maximum velocity $v_{max}$:** The motion is simple harmonic with maximum velocity $$ v_{max} = \omega A $$ where angular frequency $$ \omega = \frac{2\pi}{T} = \sqrt{\frac{k_{eq}}{m}} = \sqrt{\frac{2041.67}{13.6}} \approx \sqrt{150.12} \approx 12.25 \text{ rad/s} $$ Amplitude $A=0.044$ m, so $$ v_{max} = 12.25 \times 0.044 \approx 0.539 \text{ m/s} $$ 7. **Calculate maximum acceleration $a_{max}$:** $$ a_{max} = \omega^2 A = (12.25)^2 \times 0.044 = 150.12 \times 0.044 \approx 6.61 \text{ m/s}^2 $$ **Final answers:** - (a) Period $T \approx 0.513$ s; Frequency $f \approx 1.95$ Hz - (b) Maximum velocity $v_{max} \approx 0.539$ m/s; Maximum acceleration $a_{max} \approx 6.61$ m/s$^2$