Exploring Simple Harmonic Motion Using a Spring-Mass System

Understanding the behavior of a spring-mass system undergoing simple harmonic motion is a fundamental concept in physics and engineering. This article will delve into a specific problem where a spring with a spring constant ( k 20.0 , text{N/m} ) is attached to a mass ( m 0.750 , text{kg} ) on a frictionless table. The mass is stretched to 10.0 cm and released at time ( t 0 , text{s} ). The objective is to determine the position of the mass at a given velocity of ( 0.25 , text{m/s} ).

Problem Statement and Solution Approach

The problem involves understanding the dynamics of a spring-mass system, particularly the points where the kinetic energy (KE) and potential energy (PE) are balanced. If the system is frictionless, the conservation of energy principle plays a crucial role. According to the conservation of energy, the total energy at any point is the sum of the kinetic and potential energies. When the spring is at maximum compression or extension, the total energy is entirely potential energy.

The key equation for the conservation of energy in this system is:

KE ( ) PE Constant.

At the maximum extension, the velocity is zero, and the total energy is purely potential energy. Hence, we can write:

[ frac{1}{2}kx^2 frac{1}{2}kA^2 ]

Where ( A ) is the amplitude of the oscillation. This simplifies to:

[ x frac{A}{sqrt{2}} ]

Given the amplitude ( A 10.0 , text{cm} 0.100 , text{m} ), we can calculate:

[ x frac{0.100 , text{m}}{sqrt{2}} 0.0707 , text{m} 7.07 , text{cm} ]

From this, we see that the mass will be either stretched or compressed by 7.07 cm from the equilibrium position at the points of maximum potential energy.

Analyzing Velocity and Position in Simple Harmonic Motion

To find the position at a specific velocity, we need to use the equations of simple harmonic motion. The position ( x ) as a function of time is given by:

[ x(t) A cos(omega t) ]

Where ( omega ) (angular frequency) is defined as:

[ omega sqrt{frac{k}{m}} ]

Substituting the given values:

[ omega sqrt{frac{20.0 , text{N/m}}{0.750 , text{kg}}} sqrt{26.67 , text{s}^{-2}} approx 5.16 , text{s}^{-1} ]

The velocity ( v ) of the mass is the derivative of the position with respect to time:

[ v(t) -A omega sin(omega t) ]

Setting ( v(t) 0.25 , text{m/s} ), we solve for ( sin(omega t) ).

[ 0.25 -7.07 times 5.16 sin(omega t) ]

[ sin(omega t) -frac{0.25}{36.3} approx -0.0069 ]

Using the inverse sine function, we find:

[ omega t sin^{-1}(-0.0069) approx -0.0069 , text{radians} ]

Now, solving for time ( t ):

[ t frac{-0.0069}{5.16} approx -0.00133 , text{s} ]

We need to check if this time corresponds to when the velocity is positive. Given the periodic nature of the sine function, there will be another point in one period where the velocity is positive. The period ( T ) of the oscillation is given by:

[ T frac{2pi}{omega} frac{2pi}{5.16} approx 1.22 , text{s} ]

The position at this time can be found using:

[ x(t) 0.100 , text{m} cos(5.16 times 0.00133) 0.100 , text{m} cos(-0.0069) approx 0.100 , text{m} cos(0.0069) ]

[ x(t) approx 0.100 , text{m} times 0.999939 approx 0.09999 , text{m} -0.087 , text{m} ]

Hence, the position of the mass is approximately compressed by 8.7 cm from its equilibrium position.

Conclusion

Understanding the dynamics of a spring-mass system and applying principles of simple harmonic motion allows us to solve complex problems efficiently. By using the conservation of energy and the equations of motion, we can determine the position and velocity of the mass at any given time.

Key takeaways from this problem include:

Conservation of energy in spring-mass systems. Simple harmonic motion and its equations. Mechanics of finding positions and velocities in oscillating systems.

Feel free to explore more problems and deepen your understanding of these concepts.