15.1 Simple harmonic motion (Page 6/16)

University physics volume 1 Page 6 / 16

F_{net} = k y - k y_{0} - (k y_{0} - k y_{1}) = − k (y - y_{1}) .

Recall that $y_{1}$ is just the equilibrium position and any position can be set to be the point $y = 0.00 m .$ So let’s set $y_{1}$ to $y = 0.00 m .$ The net force then becomes

\begin{matrix} F_{net} & = & − k y; \\ m \frac{d^{2} y}{d t^{2}} & = & − k y . \end{matrix}

This is just what we found previously for a horizontally sliding mass on a spring. The constant force of gravity only served to shift the equilibrium location of the mass. Therefore, the solution should be the same form as for a block on a horizontal spring, $y (t) = A cos (ω t + ϕ) .$ The equations for the velocity and the acceleration also have the same form as for the horizontal case. Note that the inclusion of the phase shift means that the motion can actually be modeled using either a cosine or a sine function, since these two functions only differ by a phase shift.

A series of 10 illustrations of a ball, attached to a vertical spring, is shown. The illustrations are displayed next to each other, with the tops of the springs aligned. The vertical positions y = + A, y = 0, and y = -A are labeled on the right. Working our way from left to right: In the left-most drawing, the spring is compressed so the ball is at y = + A and at rest. In the second drawing, the ball is at y = 0 and is moving downward. In the third drawing, the spring is stretched so that the ball is at y = - A and at rest. In the fourth drawing, the ball is at y = 0 and is moving upward. In the fifth drawing, the spring is compressed so the ball is at y = + A and at rest. In the sixth drawing, the ball is at y = 0 and is moving downward. In the seventh drawing, the spring is stretched so that the ball is at y = - A and at rest. In the eighth drawing, the ball is at y = 0 and is moving upward. In the ninth drawing, the spring is compressed so the ball is at y = + A and at rest. In the tenth drawing, the ball is at y = 0 and is moving downward. Below these illustrations is a series of graphs, aligned vertically. The top graph is of position as a function of time. The vertical axis is position y, with a range of – A to +A. The horizontal axis is time t, labeled in increments of T. The graph has value y=+A at t=0 and oscillates two and one quarter cycles. The horizontal distance between maxima is labeled as T and the vertical distance between the horizontal axis and the maximum is labeled as amplitude A. The middle graph is of velocity as a function of time. The vertical axis is velocity v, with a range of minus v sub max to v max. The horizontal axis is time t, labeled in increments of T. The graph has value v=0 and a negative slope at t=0, and oscillates two and one quarter cycles. The bottom graph is of acceleration as a function of time. The vertical axis is acceleration a, with a range of minus a sub max to a max. The horizontal axis is time t, labeled in increments of T. The graph has value a equals minus a sub max and a and oscillates two and one quarter cycles. Below the graphs are three illustrations of the ball on the spring. The positions y = + A, y=0, and y = -A are labeled on the right. In the leftmost diagram, a hand holds the ball and the length of the spring is labeled as the unstrained length. This position is above the y = + A position. In the middle picture, the ball is not being held and is at a lower position labeled as the equilibrium position. This position is y = 0. In the rightmost diagram, the ball is shown in four different positions. These positions are y = + A, just above y = 0, just below y = 0 , and at y = -A. The spring is shown only with its bottom attached to the ball at the y = + A position. — Graphs of y ( t ), v ( t ), and a ( t ) versus t for the motion of an object on a vertical spring. The net force on the object can be described by Hooke’s law, so the object undergoes SHM. Note that the initial position has the vertical displacement at its maximum value A ; v is initially zero and then negative as the object moves down; the initial acceleration is negative, back toward the equilibrium position and becomes zero at that point.

Summary

Periodic motion is a repeating oscillation. The time for one oscillation is the period T and the number of oscillations per unit time is the frequency f . These quantities are related by $f = \frac{1}{T}$ .
Simple harmonic motion (SHM) is oscillatory motion for a system where the restoring force is proportional to the displacement and acts in the direction opposite to the displacement.
Maximum displacement is the amplitude A . The angular frequency $ω$ , period T , and frequency f of a simple harmonic oscillator are given by $ω = \sqrt{\frac{k}{m}}$ , $T = 2 π \sqrt{\frac{m}{k}}, and f = \frac{1}{2 π} \sqrt{\frac{k}{m}}$ , where m is the mass of the system and k is the force constant.
Displacement as a function of time in SHM is given by $x (t) = A cos (\frac{2 π}{T} t + ϕ) = A cos (ω t + ϕ)$ .
The velocity is given by $v (t) = − A ω sin (ω t + ϕ) = − v_{max} sin (ω t + ϕ), where v_{max} = A ω = A \sqrt{\frac{k}{m}}$ .
The acceleration is $a (t) = − A ω^{2} cos (ω t + ϕ) = − a_{max} cos (ω t + ϕ)$ , where $a_{max} = A ω^{2} = A \frac{k}{m}$ .

Conceptual questions

What conditions must be met to produce SHM?

The restoring force must be proportional to the displacement and act opposite to the direction of motion with no drag forces or friction. The frequency of oscillation does not depend on the amplitude.

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15.1 Simple harmonic motion (Page 6/16)

Summary

Conceptual questions

Problems

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