Problem 12 - College Physics

Question

\(\bullet\bullet\) A mass-spring system is in SHM in the horizontal direction. If the mass is \(0.25 \mathrm{~kg}\), the spring constant is \(12 \mathrm{~N} / \mathrm{m},\) and the amplitude is \(15 \mathrm{~cm}\) (a) what is the maximum speed of the mass, and (b) where does this occur? (c) What is the speed at a half-amplitude position?

Step-by-Step Solution

Answer

(a) Maximum speed is 1.04 m/s. (b) It occurs at the equilibrium position. (c) Speed at half-amplitude is 0.90 m/s.

Step 1: Identify the known values

We start by identifying the known values from the problem statement. The mass of the object is given as \( m = 0.25 \text{ kg} \), the spring constant is \( k = 12 \mathrm{\ N/m} \), and the amplitude of the oscillation is \( A = 15 \text{ cm} = 0.15 \text{ m} \).

Step 2: Determine the maximum speed

The maximum speed \( v_{max} \) of a mass-spring system in simple harmonic motion (SHM) can be calculated using the formula: \( v_{max} = A\omega \), where \( \omega = \sqrt{\frac{k}{m}} \) is the angular frequency. First, calculate \( \omega \):\( \omega = \sqrt{\frac{12 \text{ N/m}}{0.25 \text{ kg}}} = \sqrt{48} \approx 6.93 \text{ rad/s} \).Then, substitute \( A \) and \( \omega \) into the formula for \( v_{max} \):\( v_{max} = 0.15 \text{ m} \times 6.93 \text{ rad/s} \approx 1.04 \text{ m/s} \).

Step 3: Determine where the maximum speed occurs

In simple harmonic motion, the maximum speed occurs at the equilibrium position, where the potential energy is zero and the kinetic energy is maximum.

Step 4: Compute the speed at half-amplitude position

The speed \( v \) at a position \( x \) in SHM is given by \( v = \omega \sqrt{A^2 - x^2} \). For the half-amplitude position, \( x = \frac{A}{2} = 0.075 \text{ m} \).Substitute the values:\( v = 6.93 \text{ rad/s} \cdot \sqrt{0.15^2 - 0.075^2} \).\( = 6.93 \cdot \sqrt{0.0225 - 0.005625} \)\( = 6.93 \cdot \sqrt{0.016875} \)\( \approx 6.93 \cdot 0.12985 \approx 0.90 \text{ m/s} \).

Key Concepts

Mass-Spring System

A mass-spring system is a classic example of simple harmonic motion. It consists of a mass attached to a spring that is either compressed or extended from its natural length. When released, the mass tends to oscillate back and forth. This movement creates an ideal opportunity for studying oscillatory behavior in physics.
Key aspects of a mass-spring system:

• The mass refers to the amount of matter in the object, in our example it is 0.25 kg.
• The spring constant, noted as \( k \), measures the stiffness of the spring. A larger value means a stiffer spring. In this problem, the spring constant is 12 N/m.
• The amplitude \( A \) is the maximum extension or compression from the equilibrium position—in this exercise, 15 cm (or 0.15 m).

Understanding how these three components interact is key to analyzing the motion of the system.

Angular Frequency

Angular frequency, represented by \( \omega \), is a crucial concept used to describe oscillations. It indicates how fast an oscillation completes its cycle. In the context of a mass-spring system engaging in simple harmonic motion, angular frequency can be calculated using:

• \( \omega = \sqrt{\frac{k}{m}} \), where \( k \) is the spring constant and \( m \) is the mass.

For the exercise, substituting \( k = 12 \mathrm{\ N/m} \) and \( m = 0.25 \text{ kg} \), we find
\[ \omega = \sqrt{48} \approx 6.93 \text{ rad/s}. \]
This value tells us that the system completes approximately 6.93 radians of oscillation each second.

• Angular frequency relates directly to the velocity and energy dynamics in the system.
• A higher angular frequency suggests a faster and more energetic oscillation.

With this, we can determine parameters like maximum speed, which is vital in understanding system behavior fully.

Equilibrium Position

The equilibrium position is a central concept in studying oscillatory systems, such as a mass-spring system. It refers to the position where the forces on the mass are balanced, and there is no net force acting on it, allowing the mass to stay at rest if undisturbed.
In simple harmonic motion,

• The mass crosses this position every cycle of its oscillation.
• At this point, the potential energy of the system is zero, and kinetic energy is at its maximum.
• This is where the maximum speed of the mass is reached, as demonstrated in the solution for the exercise.

When the mass is displaced from this equilibrium position and released, it accelerates back toward it, under the influence of the restoring force exerted by the spring. This motion characterizes the simple harmonic nature of the mass-spring system, where the interplay of potential and kinetic energy defines its dynamic behavior.