Question

An ideal spring has a spring constant \(k=25 \mathrm{N} / \mathrm{m}\). The spring is suspended vertically. A 1.0 -kg body is attached to the unstretched spring and released. It then performs oscillations. (a) What is the magnitude of the acceleration of the body when the extension of the spring is a maximum? (b) What is the maximum extension of the spring?

Short answer

Answer: The magnitude of the acceleration at maximum extension is 19.6 m/s², and the maximum extension of the spring is 0.784 m.

Step-by-step solution

(a) Find the acceleration at maximum extension

To find the acceleration at maximum extension, we can use the Hooke's Law formula, which states that the force exerted by a spring is proportional to its displacement from equilibrium: \(F = -kx\) Where \(F\) is the force exerted by the spring, k is the spring constant, and x is the displacement from equilibrium. Now, using Newton's second law, we can find the acceleration (\(a\)) of the mass (\(m\)) at maximum extension: \(F = ma\) Combining Hooke's Law and Newton's second law, we have: \(-kx = ma\) Now, when the extension is maximum, the displacement is also maximum (\(x_{max}\)). So, we can rewrite and rearrange the equation to find the acceleration at maximum extension \(a_{max}\): \(a_{max} = -\frac{kx_{max}}{m}\) In this case, \(k = 25\,\text{N/m}\) and \(m = 1.0\,\text{kg}\). Since we haven't found the maximum extension yet, we will leave \(x_{max}\) in the equation.

(b) Finding the maximum extension

To find the maximum extension, we can use the conservation of mechanical energy with no friction or air resistance as follows: \(E_{total} = E_{elastic} + E_{gravitational} = \frac{1}{2} kx_{max}^2 + \frac{1}{2}m \dot{x}_{max}^2 + mgx_{max}\) In this case, the total mechanical energy is equal to the potential energy stored in the spring when the body is at its maximum extension (\(x_{max}\)): \(\frac{1}{2} kx_{max}^2 = mgx_{max}\) Divide both sides by \(x_{max}\): \(\frac{1}{2} kx_{max} = mg\) Now, we can solve for the maximum extension: \(x_{max} = \frac{2mg}{k}\) Now, plug in the given values: \(x_{max} = \frac{2(1.0\,\text{kg})(9.8\,\text{m/s}^2)}{25\,\text{N/m}}\) Calculate the maximum extension: \(x_{max} = 0.784\,\text{m}\) Now that we have the maximum extension, we can plug this back into the equation for acceleration at maximum extension: \(a_{max} = -\frac{(25\,\text{N/m})(0.784\,\text{m})}{1.0\,\text{kg}}\) Calculate the acceleration at maximum extension: \(a_{max} = -19.6\,\text{m/s}^2\) So, the magnitude of the acceleration at maximum extension is 19.6 m/s², and the maximum extension of the spring is 0.784 m.