Problem 26 - College Physics

Question

A \(0.500 \mathrm{~kg}\) 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 \(x=-0.015 \mathrm{~m},\) (c) the magnitude of the maximum acceleration of the glider, (d) the acceleration of the glider at \(x=-0.015 \mathrm{~m},\) and (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) 1.112 m/s, (c) 36 m/s², (d) 13.5 m/s², (e) 0.36 J.

Step 1: Calculate Maximum Speed

The maximum speed of an object in simple harmonic motion is given by the formula \( v_{max} = \omega A \), where \( \omega \) is the angular frequency and \( A \) is the amplitude. First, calculate \( \omega \) using the formula \( \omega = \sqrt{\frac{k}{m}} \), where \( k = 450 \; \mathrm{N/m} \) and \( m = 0.500 \; \mathrm{kg} \).\[ \omega = \sqrt{\frac{450}{0.500}} = \sqrt{900} = 30 \; \mathrm{rad/s} \]Now, use \( \omega \) to find the maximum speed:\[ v_{max} = 30 \times 0.040 = 1.2 \; \mathrm{m/s} \]

Step 2: Determine Speed at x = -0.015 m

The speed \( v \) at a position \( x \) in simple harmonic motion can be calculated using the energy conservation principle: \( v = \sqrt{\omega^2 (A^2 - x^2)} \).\[ v = \sqrt{30^2 (0.040^2 - (-0.015)^2)} = \sqrt{900 \times (0.0016 - 0.000225)} \]\[ v = \sqrt{900 \times 0.001375} = \sqrt{1.2375} \approx 1.112 \; \mathrm{m/s} \]

Step 3: Calculate Maximum Acceleration

The maximum acceleration is given by \( a_{max} = \omega^2 A \).\[ a_{max} = 30^2 \times 0.040 = 900 \times 0.040 = 36 \; \mathrm{m/s^2} \]

Step 4: Find Acceleration at x = -0.015 m

The acceleration \( a \) at a position \( x \) is found using \( a = -\omega^2 x \).\[ a = -30^2 \times (-0.015) = -900 \times (-0.015) = 13.5 \; \mathrm{m/s^2} \]

Step 5: Compute Total Mechanical Energy

The total mechanical energy in simple harmonic motion is calculated with \( E = \frac{1}{2} k A^2 \).\[ E = \frac{1}{2} \times 450 \times (0.040)^2 = \frac{1}{2} \times 450 \times 0.0016 = 0.36 \; \mathrm{J} \]

Key Concepts

Maximum Speed

In simple harmonic motion, the maximum speed occurs as the object passes through the equilibrium point. This speed can be determined using the formula:

• \( v_{max} = \omega A \)

where \( \omega \) is the angular frequency, and \( A \) is the amplitude of the motion. The angular frequency \( \omega \) is calculated by:

• \( \omega = \sqrt{\frac{k}{m}} \)

For this problem, with a spring constant \( k = 450 \; \mathrm{N/m} \) and mass \( m = 0.500 \; \mathrm{kg} \), the angular frequency \( \omega \) is \( 30 \; \mathrm{rad/s} \). Multiplying by the amplitude \( A = 0.040 \; \mathrm{m} \), we find that the maximum speed \( v_{max} \) is \( 1.2 \; \mathrm{m/s} \). This represents the fastest speed our glider achieves in its harmonic motion.

Maximum Acceleration

The maximum acceleration in simple harmonic motion is experienced at the maximum displacement from the equilibrium position. The mathematical expression for maximum acceleration is:

• \( a_{max} = \omega^2 A \)

The term \( \omega^2 \) stems from squaring the angular frequency. Given \( \omega = 30 \; \mathrm{rad/s} \) and \( A = 0.040 \; \mathrm{m} \), the maximum acceleration calculated here is \( 36 \; \mathrm{m/s^2} \). This value indicates the greatest change in velocity per unit time the glider experiences, showcasing the prioritization of forces at full stretch or compression of the spring.

Mechanical Energy

In simple harmonic motion, mechanical energy remains constant and is composed purely of kinetic and potential energies related to the motion. It can be expressed as:

• \( E = \frac{1}{2} k A^2 \)

where \( k \) is the spring constant and \( A \) is the amplitude. The glider's total mechanical energy is calculated by inserting given values, giving us:

• \( E = \frac{1}{2} \times 450 \times (0.040)^2 = 0.36 \; \mathrm{J} \)

This value of \( 0.36 \; \mathrm{J} \) represents the potential energy at maximum displacement, and it's the total energy shared between kinetic and potential forms as the glider oscillates.

Angular Frequency

Angular frequency is a key property of simple harmonic motion, relating to how rapidly an object swings back and forth. It is calculated using:

• \( \omega = \sqrt{\frac{k}{m}} \)

where \( k \) is the spring constant and \( m \) is the mass. In this scenario, \( k = 450 \; \mathrm{N/m} \) and \( m = 0.500 \; \mathrm{kg} \), leading to \( \omega = 30 \; \mathrm{rad/s} \). Angular frequency, denoted by \( \omega \), sets the rate at which cycles of motion repeat themselves and ties closely to other variables such as maximum speed and acceleration. Understanding \( \omega \) is fundamental—it links many aspects of the motion into a coherent framework.