1. **Problem statement:** An object of mass 0.5 kg is rotated in a horizontal circle by a string 1.0 m long. The maximum tension the string can withstand before breaking is 50 N. We need to find: - The greatest number of revolutions per second (frequency) the object can make. - The smallest possible time period of one revolution. 2. **Relevant formulas and concepts:** - The tension in the string provides the centripetal force: $$T = m \frac{v^2}{r}$$ - Frequency $f$ is related to velocity $v$ and radius $r$ by: $$v = 2 \pi r f$$ - Time period $T_p$ is the reciprocal of frequency: $$T_p = \frac{1}{f}$$ 3. **Step-by-step solution:** 1. Write the centripetal force equation with maximum tension: $$50 = 0.5 \times \frac{v^2}{1.0}$$ 2. Solve for $v^2$: $$v^2 = \frac{50}{0.5} = 100$$ 3. Take the square root to find $v$: $$v = \sqrt{100} = 10 \text{ m/s}$$ 4. Use the relation between velocity and frequency: $$v = 2 \pi r f \Rightarrow f = \frac{v}{2 \pi r}$$ 5. Substitute $v=10$ m/s and $r=1.0$ m: $$f = \frac{10}{2 \pi \times 1} = \frac{10}{2 \pi} = \frac{10}{6.2832} \approx 1.59 \text{ rev/s}$$ 6. Calculate the smallest time period $T_p$: $$T_p = \frac{1}{f} = \frac{1}{1.59} \approx 0.63 \text{ s}$$ 4. **Final answers:** - Greatest number of revolutions per second: $1.6$ rev/s (rounded) - Smallest possible time period: $0.625$ s (rounded)