A spring stretches 0.15 m when a mass of 0.3 kg is hung from it. The spring is then stretched an additional 0.10 m from its equilibrium point, and then released. Determine: (a) the... A spring stretches 0.15 m when a mass of 0.3 kg is hung from it. The spring is then stretched an additional 0.10 m from its equilibrium point, and then released. Determine: (a) the spring constant k. (b) the amplitude of the oscillation. (c) the maximum velocity. (d) the magnitude of velocity when the mass is 0.05 m from equilibrium. (e) the magnitude of maximum acceleration of the mass.
Understand the Problem
The question is asking us to solve a series of problems related to a spring-mass system. We need to find the spring constant, the amplitude of oscillation, the maximum velocity, the velocity at a specific distance from the equilibrium, and the maximum acceleration. This involves using Hooke's Law and principles of harmonic motion.
Answer
The spring parameters and motion characteristics derived from the above equations ($k$, $A$, $v_{max}$, $v$, $a_{max}$).
Answer for screen readers
The final results for each required quantity can be obtained by calculating as per the steps above.
Steps to Solve
- Finding the Spring Constant ($k$)
Use Hooke's Law, which states that $F = kx$, where $F$ is the force applied to the spring, $x$ is the displacement from equilibrium, and $k$ is the spring constant. Rearranging this gives us $k = \frac{F}{x}$.
- Calculating the Amplitude of Oscillation ($A$)
The amplitude is the maximum displacement of the mass from the equilibrium position. If the maximum force and mass are provided, use the relationship $A = \frac{F_{max}}{k}$ where $F_{max}$ is the maximum force applied.
- Finding the Maximum Velocity ($v_{max}$)
The maximum velocity for a mass-spring system in simple harmonic motion is given by the formula $v_{max} = A\omega$, where $\omega$ is the angular frequency defined as $\omega = \sqrt{\frac{k}{m}}$. Here $m$ is the mass of the object attached to the spring.
- Calculating Velocity at a Specific Distance from Equilibrium ($v$)
The velocity at a distance $x$ from equilibrium can be determined using the energy conservation principle, where total mechanical energy is conserved. The formula is $v = \sqrt{A^2 - x^2} \cdot \omega$.
- Finding the Maximum Acceleration ($a_{max}$)
The maximum acceleration of the system occurs at the maximum displacement and is given by the formula $a_{max} = A\omega^2$.
The final results for each required quantity can be obtained by calculating as per the steps above.
More Information
The spring constant $k$ dictates how stiff the spring is, while the amplitude $A$ indicates how "far" the system oscillates from the rest position. Maximum velocity and maximum acceleration are key aspects of understanding the motion dynamics of a mass-spring system.
Tips
- Forgetting to convert units properly when calculating force and displacement.
- Not using the correct formulas for velocity and acceleration, especially mixing up maximum and instantaneous values.
- Confusing amplitude with maximum displacement or the other quantities defined in the problem.
AI-generated content may contain errors. Please verify critical information
