Mass on a spring from equilibrium

Thread starter mancity
Start date Dec 22, 2023
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Equilibrium Formula Spring

In summary, the concept of "Mass on a spring from equilibrium" examines the behavior of a mass attached to a spring when displaced from its resting position. When the mass is pulled or pushed, it experiences a restoring force proportional to the displacement, described by Hooke's Law. The system undergoes oscillatory motion, characterized by simple harmonic motion, where the mass oscillates around the equilibrium position. Factors such as mass, spring constant, and damping influence the oscillation's period and amplitude. This model illustrates fundamental principles of mechanics and energy conservation in oscillatory systems.

Dec 22, 2023

mancity

Homework Statement: An object with mass m is suspended at rest from a spring with a spring constant of 200 N/m. The length of the spring is 5.0 cm longer than its unstretched length L, as shown above. A person then exerts a force on the object and stretches the spring an additional 5.0 cm. What is the total energy stored in the spring at the new stretch length?

Relevant Equations: Fs=1/2kx^2

Can someone explain that, when using the formula (Fs=1/2 kx^2) why do we use x=0.1m instead of 0.05m? Seems like a simple concept but why isn't it 0.05m (since 0.05m from equilibrium). Thanks.

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Dec 22, 2023

nasu

Homework Helper

If you use the formula ##E=\frac{1}{2}kx^2##, the reference for energy (zero energy) is the position with the spring unstretched. So, you need the extension of the spring relative to the unstretched position.

FAQ: Mass on a spring from equilibrium

What is the equilibrium position in a mass-spring system?

The equilibrium position in a mass-spring system is the point where the net force acting on the mass is zero. At this position, the spring is neither stretched nor compressed, and the mass remains at rest if no other forces are acting on it.

How does Hooke's Law relate to a mass-spring system?

Hooke's Law states that the force exerted by a spring is directly proportional to its displacement from the equilibrium position and is given by F = -kx, where F is the force, k is the spring constant, and x is the displacement. This law describes the restoring force that brings the mass back to equilibrium.

What is the formula for the period of oscillation for a mass-spring system?

The period of oscillation T for a mass-spring system is given by the formula T = 2π√(m/k), where m is the mass and k is the spring constant. This formula indicates that the period depends on the mass of the object and the stiffness of the spring.

What factors affect the amplitude of oscillation in a mass-spring system?

The amplitude of oscillation in a mass-spring system is primarily affected by the initial displacement from the equilibrium position and the initial velocity of the mass. External forces, such as damping or driving forces, can also affect the amplitude over time.

What is the difference between damped and undamped oscillations in a mass-spring system?

Undamped oscillations occur when there is no energy loss in the system, and the mass continues to oscillate indefinitely with constant amplitude. Damped oscillations occur when there is energy loss due to factors like friction or air resistance, causing the amplitude to decrease over time until the mass eventually comes to rest.