Physics (Mass on a Spring)
More of Hookes Law
Again, the reason to use a negative sign in the equation for Hooke's law is that the force exerted by the spring will always be in the opposite direction of the displacement. The law is best explained by example. Imagine holding a rubber band between two fingers as shown in the image. Imagine moving your fingers apart to stretch it. When you move your fingers apart, you are applying force. The more force you apply, the farther the rubber band will stretch. Now imagine relaxing your fingers. Relaxing your fingers eliminates the applied force causing the rubber band to become shorter—closer to its equilibrium position. In other words, the force to return the rubber band to its equilibrium position is proportional and in the opposite direction of how far the rubber band is stretched. Let's apply Hooke's law to this word problem. Solve for the force needed to restore a stretched spring to its equilibrium position. What is the restoring force of a spring that has been stretched 0.42 meters and has a spring constant of 85 newtons/meter? Use the formula F = -kx to solve the problem, given the following: F = force in newtons (the unknown) k = the spring constant of 85 newtons/meter x = the displacement (or stretch) of 0.42 m So, F = -(85 N/m)(0.42 m) = -36 N That means that a force of 36 newtons opposite the direction of the displacement is needed to restore the spring to its equilibrium position. (Note that F and x aren't shown here in bold as vectors because only their numerical values apply to this problem.)
Stuff
Almost every elastic object will behave according to Hooke's law. But when you stretch an elastic object beyond its limit, Hooke's law no longer applies. The spring will not return to its original shape. Perhaps you learned this truth the hard way. You may recall playing with a spring toy like the one in the image. It was fun to stretch it from one hand to the other or to make it "walk" down steps. Perhaps at one point, you wanted to stretch out the wire to see how long it was. You pulled the wire hard to straighten it. In fact, you pulled so hard that the spring exceeded its elastic limit. It could no longer follow Hooke's law and spring back to its equilibrium position.
Whatever
As you pull down on the mass the spring stretches, as shown in figure B. Notice that the mass extends past the point of equilibrium, below the white reference line. The displacement—the distance away from the resting position—has increased. The restoring force of the spring will now increase because the spring needs to balance the force from the downward pull plus the weight of the mass (mg). When you let go of the mass, it moves upward, increasing the velocity and acceleration of the mass. The velocity will reach zero when the distance from the resting point is at its greatest because the mass will stop for a moment and change direction (move up). The restoring force will actually decrease as the mass moves up closer to the resting position. You can measure the period as the time it takes for the mass to move up and down in one complete cycle.At the moment the mass reaches the equilibrium position, the restoring force and the force of the mass (mg) are equal once again and acceleration is zero. The mass will then keep moving upward, rising above the equilibrium position, shown by the top arrow in figure C. The simple harmonic motion of the mass on the spring shows that the spring will continue to move up and down until it returns to the resting position. When the spring stops moving, displacement, period, velocity, and acceleration return to zero.
Reviewing something
Let's review what happens to velocity and acceleration as the mass moves on the spring. Velocity changes and reverses direction as the mass moves from fully stretched to fully compressed. Here's what happens to velocity during periodic motion: Velocity is zero when displacement is at a maximum. Conversely, velocity is at its maximum when displacement is zero. Here's what happens to acceleration during periodic motion: When displacement is zero, acceleration is also zero because there's no net force on the mass at this position. The spring's upward tension exactly balances the gravitational force. Again conversely, when displacement is at a maximum, acceleration is also at maximum because the spring applies maximum force to restore the mass to the equilibrium position.
Motion of Mass on a Spring
Recall that the motion of a spring is classified as periodic, or simple harmonic, motion. It vibrates back and forth, or up and down, over the same path. In this part of the lesson, you'll study simple harmonic motion in greater detail. First, you'll read about the period, displacement, velocity, and acceleration of the mass on a spring. Then, you'll look at the motion of mass on a spring at different times in simple harmonic motion. Let's start with period, which is the time in seconds it takes for the spring to make one complete cycle, or one up-and-down motion. You can calculate it using this formula: T = 2pi (sqrt)m/k T represents the period, m represents the mass on the spring, and k represents the spring constant. Recall that the displacement is the distance the spring moves away from the equilibrium or resting position. Simply, it is the length of the stretch or compression of the spring.
Hookes Law
Robert Hooke (1635-1703) was a well-established English architect, philosopher, and scientist. Most of the accounts of Hooke's life were destroyed, but some information remains from an unfinished autobiography that he started in 1696. Hooke was involved in many projects and explorations. In the field of physics, he studied the elasticity of objects and how much deformation an elastic object could withstand. Hooke's law states that the force applied to a spring and the length that it stretches or compresses are directly proportional. The force required to restore a spring to its equilibrium position is called the restoring force. The equilibrium position is the resting position where no stretch or compression exists in the spring. Use this equation to find the restoring force: F = -kx. The k represents the spring constant, which is a measure of the stiffness of the spring. Its units are in newtons/meter. The x represents the displacement from its equilibrium, or resting, position. Note that the spring law typically acts in one dimension. The force (F) and the displacement (x) are in the same dimension but in opposite directions. Because the force and displacement are normally one-dimensional, you'll often see the variable x used for the displacement, even if the displacement is at some angle to the horizontal or even if it's vertical. It's important to note that the negative sign is used because the restoring force exerted by the spring will always be in the opposite direction of the displacement. The spring constant (k) determines how much force is required to restore a spring to its original position (the stiffer the spring, the greater the force required).
Omg I just need to pass
The demonstration about mass on a spring that you just studied shows these concepts: Force is dependent and directly proportional to the displacement of the mass. Restoring the force of a spring is independent of the mass of the object on the spring. If the mass is suspended, and therefore acted on by gravity, it will stretch the spring with a force equal to the weight of the mass (mg). We can calculate the period when we know the mass of the object on the spring and its spring constant, k. We measure displacement by the distance the mass moves up and down from the equilibrium position.