Question

A 0.150-kg toy is undergoing SHM on the end of a horizontal spring with force constant \(k =\) 300 N/m. When the toy is 0.0120 m from its equilibrium position, it is observed to have a speed of 0.400 m/s. What are the toy's (a) total energy at any point of its motion; (b) amplitude of motion; (c) maximum speed during its motion?

Short answer

(a) Total energy is 0.0336 J. (b) Amplitude is 0.0133 m. (c) Maximum speed is 0.664 m/s.

Step-by-step solution

Find total energy

For a system in simple harmonic motion (SHM), the total mechanical energy is the sum of kinetic and potential energy. At any point, the total energy can be calculated as \[ E = rac{1}{2} k A^2 = rac{1}{2} k x^2 + rac{1}{2} m v^2 \]where \(x = 0.0120\, \text{m}\), \(v = 0.400\, \text{m/s}\), \(m = 0.150\, \text{kg}\), and \(k = 300\, \text{N/m}\). Substituting the known values:\[E = \frac{1}{2} (300) (0.0120)^2 + \frac{1}{2} (0.150) (0.400)^2 \E = 0.0216 + 0.012 = 0.0336\, \text{J}\]

Key concepts

Spring Constant

The spring constant, denoted by \( k \), is a measure that describes how stiff or strong a spring is. It is essential in determining how the spring will react to different forces. Specifically, it is part of Hooke's Law, which states that the force exerted by a spring is directly proportional to its displacement from the equilibrium position. This relationship is expressed as: \( F = -kx \), where:

• \( F \) is the force exerted by the spring (in Newtons).
• \( k \) is the spring constant (in N/m).
• \( x \) is the displacement from the equilibrium position (in meters).

In the given exercise, the spring constant is 300 N/m. This indicates a relatively strong spring, capable of exerting a considerable force when compressed or stretched. Understanding the spring constant is critical for analyzing how the system behaves under SHM.

Total Mechanical Energy

The total mechanical energy of a system in Simple Harmonic Motion (SHM) is constant as long as no external forces (like friction) influence the motion. It comprises both kinetic energy (due to the motion of the mass) and potential energy (stored in the spring). Thus, the energy can be described by the equation:$$ E = \frac{1}{2} k A^2 = \frac{1}{2} k x^2 + \frac{1}{2} m v^2 $$

• \( A \) is the amplitude, the maximum displacement from the equilibrium position.
• \( x \) is the current displacement.
• \( v \) is the velocity of the object.
• \( m \) is the mass.

The exercise demonstrates this by calculating the total energy at a specified position. At 0.0120 m from equilibrium with a velocity of 0.400 m/s, the mechanical energy is 0.0336 J. This energy remains constant throughout the motion.

Amplitude of Motion

The amplitude of motion in SHM refers to the maximum distance the object moves from its equilibrium position. It is a crucial parameter because it dictates how far the object travels during each oscillation. The amplitude can be determined when the total mechanical energy is known, and the potential energy is at its maximum:- At the maximum displacement, the velocity is zero, and all the mechanical energy is potential energy.- The amplitude \( A \) can be found using the formula: \[ A = \sqrt{\frac{2E}{k}} \]Given the total energy calculated earlier as 0.0336 J and the spring constant \( k \) of 300 N/m, we can determine the amplitude of motion. This value signifies the greatest distance the system can reach from the central equilibrium position before returning.

Maximum Speed

In Simple Harmonic Motion, the maximum speed of the object occurs as it passes through the equilibrium position. At this point, all the mechanical energy is converted to kinetic energy, and thus the speed is at its highest. The maximum speed \( v_{max} \) can be calculated using the formula:\[ v_{max} = \sqrt{\frac{k}{m}} \cdot A \]

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

The exercise requires finding this speed based on the given values. Determine amplitude \( A \) from the total energy and spring constant, and plug these values into the equation above to find \( v_{max} \). Recognizing and understanding where the velocity reaches its peak is key to mastering the dynamics of SHM.