Problem 27 - University Physics with Modern Physics

Question

A 0.500 -kg glider, attached to the end of an ideal spring with force constant \(k=450 \mathrm{N} / \mathrm{m},\) undergoes \(\mathrm{SHM}\) 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 \(x=-0.015 \mathrm{m} ;\) (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) 1.1115 m/s; (c) 36 m/s²; (d) 13.5 m/s²; (e) 0.36 J.

Step 1: Identify Maximum Speed Formula

The maximum speed of an object in simple harmonic motion (SHM) is given by the formula \(v_{max} = A \omega\), where \(A\) is the amplitude and \(\omega\) is the angular frequency. The angular frequency \(\omega\) is calculated using the formula \(\omega = \sqrt{\frac{k}{m}}\), where \(k\) is the spring constant and \(m\) is the mass of the glider.

Step 2: Calculate Angular Frequency

Substitute the given values into the angular frequency formula: \(\omega = \sqrt{\frac{450}{0.500}} = \sqrt{900} = 30 \, \text{rad/s}\). This means the angular frequency of the glider is \(30 \, \text{rad/s}\).

Step 3: Calculate Maximum Speed

Substitute the amplitude and angular frequency into the maximum speed formula: \(v_{max} = 0.040 \, \text{m} \times 30 \, \text{rad/s} = 1.2 \, \text{m/s}\). Thus, the maximum speed of the glider is \(1.2 \, \text{m/s}\).

Step 4: Determine Speed at x = -0.015 m

The speed at a position \(x\) can be determined using the formula \(v = \omega \sqrt{A^2 - x^2}\). Substitute the values: \(v = 30 \, \text{rad/s} \times \sqrt{0.040^2 - (-0.015)^2}\). Simplifying: \(v = 30 \times \sqrt{0.0016 - 0.000225} = 30 \times \sqrt{0.001375}\). So \(v \approx 30 \times 0.03705 = 1.1115 \, \text{m/s}\).

Step 5: Calculate Maximum Acceleration

The maximum acceleration in SHM is given by \(a_{max} = A \omega^2\). Substitute the values: \(a_{max} = 0.040 \, \text{m} \times (30 \, \text{rad/s})^2 = 0.040 \times 900 = 36 \, \text{m/s}^2\). Hence, the maximum acceleration is \(36 \, \text{m/s}^2\).

Step 6: Find Acceleration at x = -0.015 m

The acceleration at any point \(x\) is given by \(a = -\omega^2 x\). Substitute the values: \(a = -(30^2) \times (-0.015) = -900 \times (-0.015) = 13.5 \, \text{m/s}^2\). Thus, the acceleration at \(x = -0.015 \, \text{m}\) is \(13.5 \, \text{m/s}^2\).

Step 7: Calculate Total Mechanical Energy

The total mechanical energy in a spring-mass system is given by \(E = \frac{1}{2} k A^2\). Substitute the given values to calculate: \(E = \frac{1}{2} \times 450 \, \text{N/m} \times (0.040 \, \text{m})^2 = 0.36 \, \text{J}\). Therefore, the total mechanical energy is \(0.36 \, \text{J}\).

Key Concepts

Maximum Speed in SHM

In Simple Harmonic Motion (SHM), an object's speed constantly changes. However, there's a point at which it reaches maximum speed. This happens when the object passes through the equilibrium point.
To find this maximum speed, we use the formula:\( v_{\text{max}} = A \omega \)where:

• \( A \) is the amplitude, or the maximum distance from the equilibrium.
• \( \omega \) is the angular frequency, which we will discuss further in another section.

For the glider in question, with an amplitude of 0.040 m and an angular frequency of 30 rad/s, the calculation is straightforward:\[ v_{\text{max}} = 0.040 \times 30 = 1.2 \text{ m/s} \]This means the glider moves fastest, at 1.2 m/s, as it passes through the center of its motion.

Angular Frequency Calculation

Angular frequency is a measure of how quickly the object oscillates back and forth. It's essential in determining various properties of SHM, including speed and energy.
The formula to calculate angular frequency \( \omega \) in a spring-mass system is:\( \omega = \sqrt{\frac{k}{m}} \)where:

• \( k \) is the spring constant, indicating the stiffness of the spring.
• \( m \) is the mass of the object.

For our glider:\( \omega = \sqrt{\frac{450}{0.500}} = \sqrt{900} = 30 \text{ rad/s} \)Thus, the angular frequency of the glider is 30 radians per second. This value helps in calculating the maximum speed and acceleration, as well as other characteristics of motion.

Mechanical Energy in Spring-Mass System

In any simple harmonic motion system, the sum of kinetic and potential energy remains constant. This total energy, known as mechanical energy, plays a crucial role in understanding the system's dynamics.
For a spring-mass system, the total mechanical energy \( E \) is calculated using the equation:\( E = \frac{1}{2} k A^2 \)where:

• \( k \) is the spring constant.
• \( A \) is the amplitude of motion.

In this scenario with a spring constant of 450 N/m and amplitude 0.040 m:\( E = \frac{1}{2} \times 450 \times (0.040)^2 \)Simplifying, we get:\( E = 0.36 \text{ J} \)
This energy indicates that despite the changes in speed and position of the glider, the total energy remains constant at 0.36 Joules throughout its motion.