Problem 28 - College Physics

Question

\(\bullet\) A 0.500 kg glider on an air track is attached to the end of an ideal spring with force constant \(450 \mathrm{N} / \mathrm{m} ;\) it undergoes simple harmonic motion with an amplitude of 0.040 \(\mathrm{m} .\) Compute (a) the maximum speed of the glider, (b) the speed of the glider when it is at \(x=-0.015 \mathrm{m},(\mathrm{c})\) the magnitude of the maximum acceleration of the glider, (d) the acceleration of the glider at \(x=-0.015 \mathrm{m},\) and \((\mathrm{e})\) the total mechanical energy of the glider at any point in its motion.

Step-by-Step Solution

Answer

(a) 1.2 m/s; (b) 1.107 m/s; (c) 36 m/s²; (d) 13.5 m/s²; (e) 0.36 J.

Step 1: Determine the natural frequency

First, we need to find the natural frequency \(\omega\) of the simple harmonic oscillator. The formula for angular frequency \(\omega\) is\[\omega = \sqrt{\frac{k}{m}}\]where \(k = 450 \, \text{N/m}\) and \(m = 0.500 \, \text{kg}\). Substituting these values gives:\[\omega = \sqrt{\frac{450}{0.500}} = 30 \, \text{rad/s} \].

Step 2: Calculate the maximum speed

The maximum speed \(v_{max}\) of the glider is given by the equation:\[v_{max} = \omega A\]where \(A = 0.040 \, \text{m}\) is the amplitude. Using \(\omega = 30 \, \text{rad/s}\), we find:\[v_{max} = 30 \times 0.040 = 1.2 \, \text{m/s}.\]

Step 3: Calculate the speed at x = -0.015 m

The speed \(v\) at any position \(x\) for a simple harmonic oscillator is given by:\[v = \omega \sqrt{A^2 - x^2}\]Substituting \(\omega = 30 \, \text{rad/s}\), \(A = 0.040 \, \text{m}\), and \(x = -0.015 \, \text{m}\), we get:\[v = 30 \sqrt{0.040^2 - (-0.015)^2} = 30 \sqrt{0.0016 - 0.000225}\]\[v = 30 \sqrt{0.001375} \approx 1.107 \text{ m/s}\].

Step 4: Find the maximum acceleration

The maximum acceleration \(a_{max}\) is given by:\[a_{max} = \omega^2 A\]Using \(\omega = 30 \, \text{rad/s}\) and \(A = 0.040 \, \text{m}\), we have:\[a_{max} = 30^2 \times 0.040 = 36 \, \text{m/s}^2.\]

Step 5: Calculate the acceleration at x = -0.015 m

The acceleration \(a\) at any position \(x\) is given by:\[a = -\omega^2 x\]Substituting \(\omega = 30 \, \text{rad/s}\) and \(x = -0.015 \, \text{m}\), we get:\[a = -(30)^2 \times (-0.015) = 13.5 \, \text{m/s}^2.\]

Step 6: Determine the total mechanical energy

The total mechanical energy \(E\) of the system at any point is constant and given by:\[E = \frac{1}{2}kA^2\]Substituting \(k = 450 \, \text{N/m}\) and \(A = 0.040 \, \text{m}\), we find:\[E = \frac{1}{2} \times 450 \times 0.040^2 = 0.36 \, \text{J}.\]

Key Concepts

Natural Frequency

In simple harmonic motion, the natural frequency represents how fast an object oscillates without external forces. It's the constant rate at which the system moves back and forth. For a system with a mass \(m\) and a spring constant \(k\), the natural frequency, often expressed as angular frequency \(\omega\), is determined by the equation:\[ \omega = \sqrt{\frac{k}{m}} \]For our given values, where the mass \(m\) of the glider is 0.500 kg, and the spring constant \(k\) is 450 N/m, we find:\[ \omega = \sqrt{\frac{450}{0.500}} = 30 \, \text{rad/s} \]This calculated natural frequency, 30 rad/s, defines the rate at which the glider will naturally oscillate on the air track when disturbed by an external force. Understanding the natural frequency helps explain the dynamic behavior of the system and its stability under different conditions.

Maximum Speed

The maximum speed in simple harmonic motion occurs when the oscillating object passes through the equilibrium position, where the energy is entirely kinetic. This speed is determined by the product of the natural frequency and the amplitude of motion. The formula to find the maximum speed \(v_{max}\) of the glider is:\[ v_{max} = \omega A \]Here, \(A\) is 0.040 m (the amplitude), and we've calculated \(\omega\) as 30 rad/s. Substituting these values gives:\[ v_{max} = 30 \times 0.040 = 1.2 \, \text{m/s} \]This result shows the fastest speed the glider achieves as it moves back and forth. In practical terms, this speed indicates how quickly the system can respond to disturbances.

Mechanical Energy

Mechanical energy in simple harmonic motion comprises two parts: potential energy stored in the spring and kinetic energy of the moving object. The total mechanical energy \(E\) is conserved and constant throughout the motion. It is calculated using:\[ E = \frac{1}{2}kA^2 \]With \(k\) being the spring constant (450 N/m) and \(A\) the amplitude (0.040 m), we find:\[ E = \frac{1}{2} \times 450 \times 0.040^2 = 0.36 \, \text{J} \]This constant energy signifies that the sum of kinetic and potential energies remains unchanged throughout. At the maximum amplitude, energy is entirely potential; at equilibrium, it's entirely kinetic. This energy conservation ensures that the system continues its oscillation perpetually (assuming no dissipative forces).

Maximum Acceleration

During oscillations, acceleration is maximized when the restoring force pulls or pushes the object to its peak displacement. The object's maximum acceleration \(a_{max}\) can be calculated using:\[ a_{max} = \omega^2 A \]Substituting \(\omega = 30 \, \text{rad/s}\) and \(A = 0.040 \, \text{m}\), we obtain:\[ a_{max} = 30^2 \times 0.040 = 36 \, \text{m/s}^2 \]This represents the highest acceleration experienced by the glider as it is furthest from its equilibrium position. Understanding this concept is crucial for analyzing how quickly the object can change its velocity.