Question

A block of mass \(1.605 \mathrm{~kg}\) is attached to a horizontal spring with spring constant \(14.55 \mathrm{~N} / \mathrm{m}\) and rests on a frictionless surface at the equilibrium position of the spring. The block is then pulled \(12.09 \mathrm{~cm}\) from the equilibrium position and released. What is the distance of the block from the equilibrium position after \(2.834 \cdot 10^{-1} \mathrm{~s} ?\)

Short answer

Answer: The distance of the block from the equilibrium position after 2.834 x 10^-1 seconds is approximately 0.0732 m.

Step-by-step solution

Identify the equation for the motion of a mass-spring system

The equation that governs the motion of a simple mass-spring system without any external force is the simple harmonic motion equation, given by: \(x(t) = A \cos(\omega t + \phi)\) where \(x(t)\) is the position of the mass at time \(t\), \(A\) is the amplitude (maximum displacement) of the motion, \(\omega\) is the angular frequency of the motion, and \(\phi\) is the phase angle.

Find the natural frequency of the system

The angular frequency \(\omega\) of the system is related to the mass \(m\) of the block and the spring constant \(k\) as follows: \(\omega = \sqrt{\frac{k}{m}}\) Using the given values of \(m = 1.605~kg\) and \(k = 14.55~N/m\), we can calculate the angular frequency. \(\omega = \sqrt{\frac{14.55}{1.605}} \approx 3.015~rad/s\)

Determine the amplitude and phase angle

Since the block is pulled \(12.09~cm\) (or \(0.1209~m\)) from the equilibrium position and released, this will be the maximum displacement (amplitude) of the motion. \(A = 0.1209~m\) Since the block is pulled away from the equilibrium position and released, the phase angle will be \(\phi = 0\).

Calculate the position of the block after the given time

Now, we have everything we need to find the position of the block after the given time, \(t = 2.834 \cdot 10^{-1}~s\). Plug in the values for \(A\), \(\omega\), and \(\phi\) into the simple harmonic motion equation: \(x(t) = 0.1209 \cdot \cos(3.015 \cdot 2.834 \cdot 10^{-1} + 0)\) Evaluate the expression to find the position of the block: \(x(t) \approx 0.0732 ~m\)

Calculate the distance from the equilibrium position

Since we have found the position of the block at time \(t\), we can find the distance from the equilibrium position by taking the absolute value of \(x(t)\): \(|x(t)| = |0.0732| = 0.0732 ~m\) So, the distance of the block from the equilibrium position after \(2.834 \cdot 10^{-1} ~s\) is approximately \(0.0732~m\).

Key concepts

Understanding Mass-Spring Systems

A mass-spring system is a classic example of simple harmonic motion (SHM), one of the most fundamental concepts in physics. At its simplest, a mass-spring system includes a mass attached to a spring. In a situation without friction and other external forces, this system will oscillate back and forth after being displaced from its equilibrium position.

The behavior of the system is quite predictable; it follows a repetitive path known as a sinusoidal wave. The time it takes for the mass to complete one full cycle of motion is referred to as the period. What makes it harmonic is that the force that restores the mass back to its equilibrium position is directly proportional to the displacement, but in the opposite direction.

Importance of Initial Conditions

Initial conditions, such as the initial displacement and velocity, ultimately dictate how the system behaves over time. For instance, if you stretch the spring by a certain distance and release it, that distance becomes the amplitude of the motion, which is the measure of the maximum displacement from the equilibrium.

When discussing improvements in solvable problems like the one outlined above, showing step-by-step how the amplitude is determined from initial conditions and connects to the simple harmonic motion can help students better understand the entire concept.

Exploring Angular Frequency

Angular frequency, often symbolized as \(\omega\), is a crucial term in studying simple harmonic motion. It tells us how quickly the mass oscillates back and forth within the mass-spring system. Consider it as an angular speed; it measures the rate of change of the phase angle (in radians) with respect to time, which reflects how many oscillations occur in a second.

Calculating Angular Frequency

The formula to calculate angular frequency is \(\omega = \sqrt{\frac{k}{m}}\), where \(k\) is the spring constant, which quantifies the stiffness of the spring, and \(m\) is the mass attached to the spring. Applying this to the given problem allows us to predict how fast the block will oscillate.

For students, angular frequency can often be a tricky concept as it doesn’t refer to a frequency in the conventional sense. Highlighting its significance in predicting the system's behavior can aid in a better grasp of the system's timing and rhythm.

Equilibrium Position: The Point of No Net Force

The equilibrium position in a mass-spring system is essentially the 'resting place' where the net force acting on the mass is zero. In other words, it is the position where the spring is neither stretched nor compressed. From this point, any displacement of the mass sets in motion the restoring force that drives simple harmonic motion.

Equilibrium and Motion

When the mass is released from any point other than the equilibrium, it will accelerate towards the equilibrium due to the spring's restoring force. Understanding where the equilibrium position lies and how it relates to the motion's amplitude is vital.

Again, in the context of exercises, pointing out the relationship between the equilibrium position and the subsequent motion after a disturbance (like pulling the block and releasing it) can highlight the nature of SHM. This concept helps students predict not only where the mass will move but also how it will continue to move through its course.