Chapter 9 Oscillations

“There is nothing so small, as to escape our inquiry; hence there is a new visible World discovered to the understanding.”

—Robert Hooke

Seventeenth-century British physicist Robert Hooke helped pave the way for simple harmonic motion. Previously, Newton’s laws of static equilibrium made it possible to show a relationship between stress and strain for complex objects. Building upon these, he developed laws on simple harmonic motion that are named after him today—Hooke’s Law. Hooke is credited as the discoverer of the concise mathematical relationship of a spring.

In this section, we will focus on periodic motion that’s straightforward which will help to describe oscillations and other simple harmonic motions.

SIMPLE HARMONIC MOTION

In the series of diagrams below is a fixed block on the left side of a wall. When the spring is neither stretched nor compressed (when it sits at its natural length), it is said to be in its equilibrium position. When the block is in equilibrium, the net force on the block is zero. We label this position as x = 0.

Let’s start with a spring at rest (Diagram 1). First, we pull the block to the right, where it will experience a force pulling back toward equilibrium (Diagram 2). This force brings the block back through the equilibrium position (Diagram 3), and the block’s momentum carries it past that point to location x = ‒A. At this point, the block will again be experiencing a force that pushes it toward the equilibrium position (Diagram 4). Once again, the block passes through the equilibrium position, but it’s traveling to the right this time (Diagram 5). If this is taking place in ideal conditions (no friction), this back-and-forth motion will continue indefinitely and the block will oscillate from these positions in the same amount of time. The oscillations of the block at the end of this spring provide us with the physical example of simple harmonic motion (often abbreviated SHM).

The Dynamics of SHM

Force

Since the block is accelerating and decelerating, there must be some force that is making it do so. In the case of a spring, the spring exerts a force on the block that is proportional to its displacement from its equilibrium point. Setting the equilibrium point as x = 0, the force exerted by the spring is

This is known as Hooke’s Law. The proportionality constant, k, is called the spring constant and tells us how strong the spring is. The greater the value of k, the stiffer and stronger the spring is. The minus sign in Hooke’s Law tells us that the force is a restoring force. A restoring force simply means that the force wants to return the object back to its equilibrium position. Hence, in the previous diagrams, when the block was on the left (when the position is negative), the force exerted was to the right; and when the block was on the right (when the position is positive), the force exerted was to the left. In all cases of the extreme left or right, the spring has a tendency to return to its original length or equilibrium position. This force helps to maintain the oscillations.

Solution. The displacement of the spring has a magnitude of 14 ‒ 12 = 2 cm = 0.02 m so, according to Hooke’s Law, the spring exerts a force of magnitude F = kx = (400 N/m)(0.02 m) = 8 N. Therefore, we’d have to exert this much force to keep the spring in this stretched state.

During the oscillation, the force on the block is zero when the block is at equilibrium (the point we designate as x = 0). This is because Hooke’s Law says that the strength of the spring’s restoring force is given by the equation F = kx, so F = 0 at equilibrium. The acceleration of the block is also equal to zero at x = 0, since F = 0 at x = 0 and a = F/m. At the endpoints of the oscillating region, where the block’s displacement, x, has the greatest magnitude, the restoring force and the magnitude of the acceleration are both at their maximums.

Amplitude

The maximum displacement from equilibrium is called the amplitude of oscillation and is denoted by A. So instead of writing x = x_max, we write x = A (x = ‒x_max is x = ‒A). This number tells us how far to the left and right of equilibrium the block will travel.

SHM in Terms of Energy

Another way to describe the block’s motion is in terms of energy transfers. A stretched or compressed spring stores elastic potential energy, which is transformed into kinetic energy (and back again); this shuttling of energy between potential and kinetic causes the oscillations. For a spring with spring constant k, the elastic potential energy it possesses—relative to its equilibrium position—is given by the following equation:

Notice that the farther you stretch or compress a spring, the more work you have to do, and, as a result, the more potential energy that’s stored.

In terms of energy transfers, we can describe the block’s oscillations as follows. When you initially pull the block out, you increase the elastic potential energy of the system. Upon releasing the block, this potential energy turns into kinetic energy, and the block moves. As it passes through equilibrium, U_S = 0, all the energy is kinetic. Then, as the block continues through equilibrium, it compresses the spring and the kinetic energy is transformed back into elastic potential energy.

By Conservation of Mechanical Energy, the sum K + U_S is a constant. Therefore, when the block reaches the endpoints of the oscillation region (that is, when x = ±x_max), U_S is maximized, so K must be minimized; in fact, K = 0 at the endpoints. As the block is passing through equilibrium, x = 0, so U_S = 0 and K is maximized.

Solution. First, let’s get an expression for the maximum elastic potential energy of the system:

When all this energy has been transformed into kinetic energy—which, as we discussed earlier, occurs just as the block is passing through equilibrium—the block will have a maximum kinetic energy and maximum speed of

Solution. The total energy of the oscillator is the sum of its kinetic and potential energies. By Conservation of Mechanical Energy, the sum K + U_S is a constant, so if we can determine what this sum is at some point in the oscillation region, we’ll know the sum at every point. When the block is at its amplitude position, x = 8 cm, its speed is zero; so at this position, E is easy to figure out:

This gives the total energy of the oscillator at every position. At any position x, we have

Solution. The block will come to rest when all of its initial kinetic energy has been transformed into the spring’s potential energy. At this point, the block is at its maximum displacement from equilibrium, at one of its amplitude positions, and

Period and Frequency

As you watch the block oscillate, you should notice that it repeats each cycle of oscillation in the same amount of time. A cycle is a round-trip: for example, from position x = A over to x = ‒A and back again to x = A. The amount of time it takes to complete a cycle is called the period of the oscillations, or T. If T is short, the block is oscillating rapidly, and if T is long, the block is oscillating slowly.

Another way of indicating the rapidity of the oscillations is to count the number of cycles that can be completed in a given time interval; the more completed cycles, the more rapid the oscillations. The number of cycles that can be completed per unit time is called the frequency of the oscillations, or f, and frequency is expressed in cycles per second. One cycle per second is one hertz (abbreviated Hz). Do not confuse lowercase f (frequency) with uppercase F (Force).

One of the most basic equations of oscillatory motion expresses the fact that the period and frequency are reciprocals of each other:

Solution. The period is defined as the time required for one full cycle. Moving from one end of the oscillation region to the other is only half a cycle. Therefore, if the block moves from its position of maximum spring stretch to maximum spring compression in 0.25 s, the time required for a full cycle is twice as much; T = 0.5 s. Because frequency is the reciprocal of period, the frequency of the oscillations is f = 1/T = 1/(0.5 s) = 2 Hz.

Solution. The frequency of the oscillations, in hertz (which is the number of cycles per second), is

One of the defining properties of the spring-block oscillator is that the frequency and period can be determined from the mass of the block and the force constant of the spring. The equations are as follows:

Let’s analyze these equations. Suppose we had a small mass on a very stiff spring; then intuitively, we would expect that this strong spring would make the small mass oscillate rapidly, with high frequency and short period. Both of these predictions are substantiated by the equations above, because if m is small and k is large, then the ratio k/m is large (high frequency) and the ratio m/k is small (short period).

Notice that f ≈ 2 Hz and T ≈ 0.5 s, and that these values satisfy the basic equation T = 1/f.

Solution. Since the same spring is used, k remains the same. According to the equation given on the previous page, f is inversely proportional to the square root of the mass of the block:

. Therefore, if m decreases by a factor of 4, then f increases by a factor of

= 2.

The equations we saw on the previous page for the frequency and period of the spring-block oscillator do not contain A, the amplitude of the motion. In simple harmonic motion, both the frequency and the period are independent of the amplitude. The reason that the frequency and period of the spring-block oscillator are independent of amplitude is that F, the strength of the restoring force, is proportional to x, the displacement from equilibrium, as given by Hooke’s Law: F_S = ‒kx.

Example 9 A student performs an experiment with a spring-block simple harmonic oscillator. In the first trial, the amplitude of the oscillations is 3.0 cm, while in the second trial (using the same spring and block), the amplitude of the oscillations is 6.0 cm. Compare the values of the period, frequency, and maximum speed of the block between these two trials.

Solution. If the system exhibits simple harmonic motion, then the period and frequency are independent of amplitude. This is because the same spring and block were used in the two trials, so the period and frequency will have the same values in the second trial as they had in the first. But the maximum speed of the block will be greater in the second trial than in the first. Since the amplitude is greater in the second trial, the system possesses more total energy (E =

kA²). So when the block is passing through equilibrium (its position of greatest speed), the second system has more energy to convert to kinetic, meaning that the block will have a greater speed. In fact, from Example 2, we know that v_max =

so, since A is twice as great in the second trial than in the first, v_max will be twice as great in the second trial than in the first.

Example 10 For each of the following arrangements of two springs, determine the effective spring constant, k_eff. This is the force constant of a single spring that would produce the same force on the block as the pair of springs shown in each case.

(d) Determine k_eff in each of these cases if k₁ = k₂ = k.

(a) Imagine that the block was displaced a distance x to the right of its equilibrium position. Then the force exerted by the first spring would be F₁ = ‒k₁x and the force exerted by the second spring would be F₂ = ‒k₂x. The net force exerted by the springs would be

F₁ + F₂ = ‒k₁x + (‒k₂x) = ‒(k₁ + k₂)x

Since F_eff = ‒(k₁ + k₂)x, we see that k_eff = k₁ + k₂.

(b) Imagine that the block was displaced a distance x to the right of its equilibrium position. Then the force exerted by the first spring would be F₁ = ‒k₁x and the force exerted by the second spring would be F₂ = ‒k₂x. The net force exerted by the springs would be

F₁ + F₂ = ‒k₁x + ‒k₂x = ‒(k₁ + k₂)x

As in part (a), we see that, since F_eff = ‒(k₁ + k₂)x, we get k_eff = k₁ + k₂.

(c) Imagine that the block was displaced a distance x to the right of its equilibrium position. Let x₁ be the distance that the first spring is stretched, and let x₂ be the distance that the second spring is stretched. Then x = x₁ + x₂. But x₁ = ‒F/k₁ and x₂ = ‒F/k₂, so

Therefore,

(d) If the two springs have the same force constant, that is, if k₁ = k₂ = k, then in the first two cases, the pairs of springs are equivalent to one spring that has twice their force constant: k_eff = k₁ + k₂ = k + k = 2k. In (c), the pair of springs is equivalent to a single spring with half their force constant:

Spring-Block Summary

Fortunately, this same table applies to springs, pendulums, and waves. All simple harmonic motion follows this cycle.

THE SPRING-BLOCK OSCILLATOR: VERTICAL MOTION

So far we’ve looked at a block sliding back and forth on a horizontal table, but the block could also oscillate vertically. The only difference would be that gravity would cause the block to move downward, to an equilibrium position at which, in contrast with the horizontal SHM we’ve examined, the spring would not be at its natural length. Of course, in calculating energy, the gravitational potential energy (mgh) must be included.

Consider a spring of negligible mass hanging from a stationary support. A block of mass m is attached to its end and allowed to come to rest, stretching the spring a distance d. At this point, the block is in equilibrium; the upward force of the spring is balanced by the downward force of gravity. Therefore,

Once you have solved for the distance d, simply set this position as your new equilibrium position x = 0. At this point, you can treat the vertical spring exactly as you would a horizontal spring. Just be sure to measure distances relative to the new equilibrium when solving for any values (spring force, potential energy, and so on) that have x in their formulas.

New Equilibrium

Normally our equilibrium for the spring (without a block) would be at the position y. When we attach the block, our new equilibrium point sits at (d + y). The only time you need to worry about this is if a question asks about the total length of the spring at a given moment. Use your horizontal spring equations to solve the question normally, and then simply add d.

Example 11 A block of mass m = 1.5 kg is attached to the end of a vertical spring of force constant k = 300 N/m. After the block comes to rest, it is pulled down a distance of 2.0 cm and released.

(a) What is the frequency of the resulting oscillations?

(b) What are the minimum and maximum amounts of stretch of the spring during the oscillations of the block?

(a) The frequency is given by

(b) Before the block is pulled down, to begin the oscillations, it stretches the spring by a distance calculated as follows:

Since the amplitude of the motion is 2.0 cm, the spring is stretched a maximum of 5 cm + 2.0 cm = 7 cm when the block is at the lowest position in its cycle, and a minimum of 5 cm ‒ 2.0 cm = 3 cm when the block is at its highest position.

PENDULUMS

A simple pendulum consists of a weight of mass m attached to a string or a massless rod that swings, without friction, about the vertical equilibrium position. The restoring force is provided by gravity and, as the figure on the next page shows, the magnitude of the restoring force when the bob is θ to an angle to the vertical is given by the equation:

Although the displacement of the pendulum is measured by the angle that it makes with the vertical, rather than by its linear distance from the equilibrium position (as was the case for the spring-block oscillator), the simple pendulum shares many of the important features of the spring-block oscillator. For example,

Despite these similarities, there is one important difference. Simple harmonic motion results from a restoring force that has a strength that is proportional to the displacement. The magnitude of the restoring force on a pendulum is mg sin θ, which is not proportional to the displacement (θL, the arc length, with the angle measured in radians). Strictly speaking, the motion of a simple pendulum is not really simple harmonic. However, if θ is small, then sin θ ≈ θ (measured in radians) so, in this case, the magnitude of the restoring force is approximately mgθ, which is proportional to θ. So if θ_max is small, the motion can be treated as simple harmonic.

If the restoring force is given by mgθ, rather than mg sin θ, then the frequency and period of the oscillations depend only on the length of the pendulum and the value of the gravitational acceleration, according to the following equations:

Note that neither frequency nor the period depends on the amplitude (the maximum angular displacement, θ_max); this is a characteristic feature of simple harmonic motion. Also notice that neither the frequency nor the period depends on the mass of the weight.

Solution. The equation T =

shows that T is inversely proportional to

, so if g decreases by a factor of 6, then T increases by factor of

. That is,

Chapter 9 Review Questions

Section I: Multiple Choice

1. Which of the following are characteristics of simple harmonic motion? Select two answers.

(A) The acceleration is constant.

(B) The restoring force is proportional to the displacement.

(D) The period is dependent on the amplitude.

2. A block attached to an ideal spring undergoes simple harmonic motion. The acceleration of the block has its maximum magnitude at the point where

(A) the speed is the maximum

(B) the speed is the minimum

(D) the kinetic energy is the maximum

3. A block attached to an ideal spring undergoes simple harmonic motion about its equilibrium position (x = 0) with amplitude A. What fraction of the total energy is in the form of kinetic energy when the block is at position x = A ?

(A)

(B)

(C)

(D)

4. A student measures the maximum speed of a block undergoing simple harmonic oscillations of amplitude A on the end of an ideal spring. If the block is replaced by one with twice its mass but the amplitude of its oscillations remains the same, then the maximum speed of the block will

(A) decrease by a factor of 4

(B) decrease by a factor of 2

(D) increase by a factor of 2

5. A spring-block simple harmonic oscillator is set up so that the oscillations are vertical. The period of the motion is T. If the spring and block are taken to the surface of the Moon, where the gravitational acceleration is 1/6 of its value here, then the vertical oscillations will have a period of

(A)

(B)

(C)

(D) T

6. A linear spring of force constant k is used in a physics lab experiment. A block of mass m is attached to the spring and the resulting frequency, f, of the simple harmonic oscillations is measured. Blocks of various masses are used in different trials, and in each case, the corresponding frequency is measured and recorded. If f² is plotted versus 1/m, the graph will be a straight line with slope

(A)

(B)

(D)

7. A simple pendulum swings about the vertical equilibrium position with a maximum angular displacement of 5° and period T. If the same pendulum is given a maximum angular displacement of 10°, then which of the following best gives the period of the oscillations?

(A)

(B)

(D) 2T

8. A block with a mass of 20 kg is attached to a spring with a force constant k = 50 N/m. What is the magnitude of the acceleration of the block when the spring is stretched 4 m from its equilibrium position?

(A) 4 m/s²

(B) 6 m/s²

(D) 10 m/s²

9. A block with a mass of 10 kg connected to a spring oscillates back and forth with an amplitude of 2 m. What is the approximate period of the block if it has a speed of 4 m/s when it passes through its equilibrium point?

(A) 1 s

(B) 3 s

(D) 12 s

10. A block with a mass of 4 kg is attached to a spring on the wall that oscillates back and forth with a frequency of 4 Hz and an amplitude of 3 m. What would the frequency be if the block were replaced by one with one-fourth the mass and the amplitude of the block is increased to 9 m ?

(A) 4 Hz

(B) 8 Hz

(D) 24 Hz

Section II: Free Response

1. The figure below shows a block of mass m (Block 1) that is attached to one end of an ideal spring of force constant k and natural length L. The block is pushed so that it compresses the spring to 3/4 of its natural length and is then released from rest. Just as the spring has extended to its natural length L, the attached block collides with another block (also of mass m) at rest on the edge of the frictionless table. When Block 1 collides with Block 2, half of its kinetic energy is lost to heat; the other half of Block 1’s kinetic energy at impact is divided between Block 1 and Block 2. The collision sends Block 2 over the edge of the table, where it falls a vertical distance H, landing at a horizontal distance R from the edge.

(a) What is the acceleration of Block 1 at the moment it’s released from rest from its initial position? Write your answer in terms of k, L, and m.

(b) If v₁ is the velocity of Block 1 just before impact, show that the velocity of Block 1 just after impact is v₁.

(c) Determine the amplitude of the oscillations of Block 1 after Block 2 has left the table. Write your answer in terms of L only.

(d) Determine the period of the oscillations of Block 1 after the collision, writing your answer in terms of T₀, the period of the oscillations that Block 1 would have had if it did not collide with Block 2.

(e) Find an expression for R in terms of H, k, L, m, and g.

2. A bullet of mass m is fired from a non-lethal pellet gun horizontally with speed v into a block of mass M initially at rest, at the end of an ideal spring on a frictionless table. At the moment the bullet hits, the spring is at its natural length, L. The bullet becomes embedded in the block, and simple harmonic oscillations result.

(a) Determine the speed of the block immediately after the impact by the bullet.

(b) Determine the amplitude of the resulting oscillations of the block.

3. A block of mass M oscillating with amplitude A on a frictionless horizontal table is connected to an ideal spring of force constant k. The period of its oscillations is T. At the moment when the block is at position x = A and moving to the right, a ball of clay of mass m dropped from above lands on the block.

(a) What is the velocity of the block just before the clay hits?

(b) What is the velocity of the block just after the clay hits?

(d) What is the new amplitude of the oscillations? Write your answer in terms of A, k, M, and m.

(e) Would the answer to part (c) be different if the clay had landed on the block when it was at a different position? Support your answer briefly.

(f) Would the answer to part (d) be different if the clay had landed on the block when it was at a different position? Support your answer briefly.