To solve this problem, we need to find the extension in the spring when the body of mass 200 g moves in a circular path with constant angular speed. We'll use the concepts of circular motion and Hooke's Law for this.
- First, convert the mass of the body from grams to kilograms for consistency in SI units: \(m = 200 \, \text{g} = 0.2 \, \text{kg}\)
- The spring constant is given as \(k = 12.5 \, \text{N/m}\).
- The angular speed of the body is \(\omega = 5 \, \text{rad/s}\).
- The centripetal force \(F_c\) needed for circular motion is provided by the extension in the spring and is given by: \(F_c = m \cdot \omega^2 \cdot r\)
- According to Hooke's Law, the force exerted by the spring is \(F = k \cdot x\), where \(x\) is the extension in the spring.
- Equating the forces (as the spring force provides the centripetal force): \(k \cdot x = m \cdot \omega^2 \cdot r\)
- We can express the radius \(r\) in terms of natural length \(l\) and extension \(x\) as: \(r = l + x\)
- Substituting \(r\) into the force equation gives us: \(k \cdot x = m \cdot \omega^2 \cdot (l + x)\)
- Rearranging terms to find the ratio of the extension to the natural length: \(x = \frac{m \cdot \omega^2 \cdot l}{k - m \cdot \omega^2}\)
- Substitute the values: \(x = \frac{0.2 \cdot 5^2 \cdot l}{12.5 - 0.2 \cdot 5^2}\)
- Simplify the equation: \(x = \frac{0.2 \cdot 25 \cdot l}{12.5 - 0.2 \cdot 25}\) \(x = \frac{5 \cdot l}{12.5 - 5}\) \(x = \frac{5 \cdot l}{7.5}\) \(x = \frac{2 \cdot l}{3}\)
- Thus, the ratio of the extension in the spring to its natural length is \(\frac{2}{3}\).
The correct answer is therefore \(\frac{2}{3}\), which matches with the option given as \(2:3\).