Problem 30 - University Physics with Modern Physics

Question

A 0.150 -kg toy is undergoing SHM on the end of a horizontal spring with force constant \(k=300 \mathrm{N} / \mathrm{m} .\) When the object is 0.0120 \(\mathrm{m}\) from its equilibrium position, it is observed to have a speed of 0.300 \(\mathrm{m} / \mathrm{s} .\) What are (a) the total energy of the object at any point of its motion; (b) the amplitude of the motion; \((\mathrm{c})\) the maximum speed attained by the object during its motion?

Step-by-Step Solution

Answer

(a) 0.02835 J; (b) 0.0137 m; (c) 0.615 m/s.

Step 1: Calculate the Total Mechanical Energy

The total mechanical energy of a mass-spring system undergoing simple harmonic motion (SHM) is constant and is a sum of the kinetic energy (KE) and potential energy (PE) at any point of its motion.Kinetic energy when the toy is 0.0120 m from equilibrium:\[KE = \frac{1}{2} m v^2 = \frac{1}{2} \times 0.150 \times (0.300)^2 = 0.00675 \text{ J}\]Potential energy at the same point:\[PE = \frac{1}{2} k x^2 = \frac{1}{2} \times 300 \times (0.0120)^2 = 0.0216 \text{ J}\]Total Energy:\[E = KE + PE = 0.00675 + 0.0216 = 0.02835 \text{ J}\]

Step 2: Identify the Amplitude of the Motion

The total energy of the system is also expressed in terms of the amplitude (A) of the motion, since at the amplitude, all energy is potential:\[E = \frac{1}{2} k A^2\]Setting this equal to the total energy calculated:\[0.02835 = \frac{1}{2} \times 300 \times A^2\]Solving for A:\[A^2 = \frac{2 \times 0.02835}{300}\]\[A^2 = 0.000189\]\[A = \sqrt{0.000189} \approx 0.0137 \text{ m}\]

Step 3: Calculate the Maximum Speed

The maximum speed occurs when the spring's potential energy is zero and all of the system's energy is kinetic. The maximum speed (\(v_\text{max}\)) is given by:\[v_\text{max} = \sqrt{\frac{2E}{m}}\]Substituting in the energy and mass:\[v_\text{max} = \sqrt{\frac{2 \times 0.02835}{0.150}} = \sqrt{0.378} \approx 0.615 \text{ m/s}\]

Key Concepts

Total Mechanical Energy

The total mechanical energy in a simple harmonic motion (SHM) system, such as a mass attached to a spring, is a fundamental concept. It is the sum of kinetic energy (energy associated with motion) and potential energy (energy stored in the spring due to its position) at any given point in the cycle. This energy remains constant throughout the motion, as the energy transforms back and forth between kinetic and potential.

In the given example, the toy with a mass of 0.150 kg has kinetic energy given by the formula:

• \( KE = \frac{1}{2} m v^2 \)

where \( m = 0.150 \text{ kg} \) and \( v = 0.300 \text{ m/s} \).

Calculating this yields \( KE = 0.00675 \text{ J} \). The potential energy can be calculated:

• \( PE = \frac{1}{2} k x^2 \)

where \( k = 300 \text{ N/m} \) and \( x = 0.0120 \text{ m} \). Solving gives \( PE = 0.0216 \text{ J} \).

Thus, the total mechanical energy \( E \) for the system is:

• \( E = KE + PE = 0.00675 + 0.0216 = 0.02835 \text{ J} \)

This constant energy exists through the entire motion, allowing for calculations regarding amplitude and maximum speed.

Amplitude of Motion

In SHM, the amplitude of motion refers to the maximum extent of displacement from the equilibrium position. It is a crucial measure that defines the oscillation’s range. At this peak displacement, all the system’s mechanical energy is stored as potential energy.

Given the total mechanical energy \( E = 0.02835 \text{ J} \), it can be seen how amplitude (\( A \)) ties in. When displacement is at its maximum (, the potential energy of the system equals total energy. Therefore, using

• \( E = \frac{1}{2} k A^2 \)

it can be reorganized to find \( A \).

Solving gives:

• \( A^2 = \frac{2 imes 0.02835}{300} \)
• \( A^2 = 0.000189 \)
• \( A = \sqrt{0.000189} \approx 0.0137 \, \text{m} \)

This calculated amplitude tells us the farthest point the toy will reach from its equilibrium, both to the right or left.

Maximum Speed in SHM

In the journey of simple harmonic motion, the object attains its maximum speed when the entire system’s mechanical energy is in the form of kinetic energy. This occurs at the equilibrium point where potential energy reaches zero.

The expression for maximum speed \( v_\text{max} \) stems from equating total mechanical energy to kinetic energy as:

• \( v_\text{max} = \sqrt{\frac{2E}{m}} \)

For the given toy:

• \( v_\text{max} = \sqrt{\frac{2 \times 0.02835}{0.150}} \)
• \( v_\text{max} = \sqrt{0.378} \)
• \( v_\text{max} \approx 0.615 \, \text{m/s} \)

This maximum speed point is where the toy will quickly sweep through the equilibrium, rapidly converting potential energy to kinetic. Understanding the velocity gives insight into how dynamic and swift the object’s oscillations are within the spring’s domain.