Problem 37 - College Physics

Question

\(\bullet\) \(\bullet\) One end of a stretched ideal spring is attached to an airtrack and the other is attached to a glider with a mass of 0.355 \(\mathrm{kg}\) . The glider is released and allowed to oscillate in SHM. If the distance of the glider from the fixed end of the spring varies between 1.80 \(\mathrm{m}\) and \(1.06 \mathrm{m},\) and the period of the oscillation is \(2.15 \mathrm{s},\) find (a) the force constant of the spring, (b) the maximum speed of the glider, and (c) the magnitude of the maximum acceleration of the glider.

Step-by-Step Solution

Answer

(a) The force constant is approximately 3.02 N/m. (b) The maximum speed is 1.08 m/s. (c) The maximum acceleration is 3.16 m/s².

Step 1: Calculate Amplitude

To find the amplitude of oscillation, subtract the minimum position from the maximum position and divide by 2. The amplitude is given by \( A = \frac{1.80 \ \text{m} - 1.06 \ \text{m}}{2} = 0.37 \ \text{m} \).

Step 2: Determine Angular Frequency

The angular frequency \( \omega \) is related to the period \( T \) by the formula \( \omega = \frac{2\pi}{T} \). Substitute \( T = 2.15 \ \text{s} \): \( \omega = \frac{2\pi}{2.15} \approx 2.92 \ \text{rad/s} \).

Step 3: Calculate Force Constant

The force constant \( k \) of the spring can be found using the formula \( \omega = \sqrt{\frac{k}{m}} \). Rearranging gives \( k = m\omega^2 \). Substituting \( m = 0.355\ \text{kg} \) and \( \omega = 2.92 \ \text{rad/s} \), \( k = 0.355 \cdot (2.92)^2 \approx 3.02 \ \text{N/m} \).

Step 4: Calculate Maximum Speed

The maximum speed \( v_{\text{max}} \) occurs at the equilibrium position and is given by \( v_{\text{max}} = A\omega \). Substitute \( A = 0.37 \ \text{m} \) and \( \omega = 2.92 \ \text{rad/s} \): \( v_{\text{max}} = 0.37 \times 2.92 \approx 1.08 \ \text{m/s} \).

Step 5: Calculate Maximum Acceleration

The maximum acceleration \( a_{\text{max}} \) is given by \( a_{\text{max}} = A\omega^2 \). Using \( A = 0.37 \ \text{m} \) and \( \omega = 2.92 \ \text{rad/s} \), calculate \( a_{\text{max}} = 0.37 \times (2.92)^2 \approx 3.16 \ \text{m/s^2} \).

Key Concepts

Spring Constant

The spring constant, often symbolized as \( k \), is a measure of the stiffness of a spring. It's a central concept in physics when dealing with elastic objects like springs. The spring constant tells us how much force is needed to stretch or compress the spring by a unit length. The formula \( F = kx \) describes Hooke's Law, where \( F \) is the force applied, \( k \) is the spring constant, and \( x \) is the deformation of the spring from its equilibrium position.
This property is vital when studying simple harmonic motion (SHM), as it can influence the oscillation style and energy of the system. To find the spring constant in this exercise, we use the relation from SHM given by \( \omega = \sqrt{\frac{k}{m}} \). Rearranging for \( k \), we have \( k = m\omega^2 \). With mass \( m \) and angular frequency \( \omega \) calculated, you can solve for \( k \).
The higher the spring constant, the stiffer the spring, meaning it requires more force to achieve the same deformation compared to a spring with a lower spring constant.

Angular Frequency

Angular frequency, denoted as \( \omega \), is a key parameter in the study of oscillatory systems. It represents how quickly an object oscillates in radians per second. Mathematically, it's related to the period \( T \) of the oscillation by \( \omega = \frac{2\pi}{T} \). This relationship highlights that the angular frequency decreases with increasing period, meaning slower oscillations.
In simple harmonic motion, angular frequency is crucial for determining other dynamic properties such as maximum speed and acceleration. For the given problem, determining \( \omega \) allowed us to calculate other quantities like the spring constant \( k \) and maximum acceleration \( a_{\text{max}} \).
Angular frequency is often preferred in physics because it aligns with the circular functions (sine and cosine) used to describe SHM, providing a seamless connection between linear and rotational systems.

Maximum Acceleration

Maximum acceleration in the context of simple harmonic motion occurs at the extreme points of oscillation. It can be calculated using the formula \( a_{\text{max}} = A\omega^2 \), where \( A \) is the amplitude and \( \omega \) is the angular frequency.
Acceleration in SHM aligns with the net force acting on the object as described by Newton's second law. In a spring-mass system, the maximum force, and thus maximum acceleration, happens at the endpoints of the oscillation, where speed is momentarily zero but directional change is imminent.
This detail explains why simple harmonic oscillators like pendulums or masses on springs slow and reverse direction smoothly at the amplitude limits. Understanding maximum acceleration aids in comprehending the forces and energy dynamics within oscillatory systems.

Amplitude of Oscillation

Amplitude, denoted by \( A \), is the peak value of displacement from the equilibrium position in an oscillatory motion. It's one of the simplest yet most important properties of SHM. The amplitude dictates the range of motion and the maximum potential energy of the oscillator.
In the exercise, the amplitude of oscillation is derived by subtracting the minimum position from the maximum position and halving the result. This gives a measure of how far the glider moves from the center point to either extreme during its motion.
The amplitude directly influences other characteristics of the motion, such as maximum speed and acceleration. Higher amplitudes involve greater extremes of motion, leading to increased energy requirements to sustain the motion. Understanding amplitude helps in visualizing the complete extent of the harmonics involved.