Ch10 Simple Harmonic Motion and Elasticity

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Transcript Ch10 Simple Harmonic Motion and Elasticity

Ch10. Simple Harmonic Motion and Elasticity The Ideal Spring and Simple Harmonic Motion

Applied



The constant

is called the

spring constant

A spring that behaves

F

Applied



said to be an

ideal spring

Example 1.

A Tire Pressure Gauge

In a tire pressure gauge, the air in the tire pushes against a plunger attached to a spring when the gauge is pressed against the tire valve. Suppose the spring constant of the spring is k = 320 N/m and the bar indicator of the gauge extends 2.0 cm when the gauge is pressed against the tire valve. What force does the air in the tire apply to the spring?

Conceptual Example 2.

Are Shorter Springs Stiffer Springs?

A 10-coil spring that has a spring constant k. If this spring is cut in half, so there are two 5-coil springs, what is the spring constant of each of the smaller springs?

Shorter springs are stiffer springs.

Sometimes the spring constant k is referred to as the

stiffness

of the spring, because a large value for k means the spring is “stiff,” in the sense that a large force is required to stretch or compress it.

HOOKE’S LAW RESTORING FORCE OF AN IDEAL SPRING* The restoring force of an ideal spring is (10.2) where k is the spring constant and x is the displacement of the spring from its unstrained length. The minus sign indicates that the restoring force always points in a direction opposite to the displacement of the spring.

When the restoring force has the mathematical form given by F = –kx, the type of friction-free motion illustrated in the figure is designated as “ simple harmonic motion .”

The maximum excursion from equilibrium is the amplitude A of the motion. The shape of this graph is characteristic of simple harmonic motion and is called “

sinusoidal

,” because it has the shape of a trigonometric sine or cosine function.

Check Your Understanding 1

A 0.42-kg block is attached to the end of a horizontal ideal spring and rests on a frictionless surface. The block is pulled so that the spring stretches by 2.1 cm relative to its unstrained length. When the block is released, it moves with an acceleration of 9.0 m/s 2 . What is the spring constant of the spring?

180 N/m

2.1cm

kx = ma

 2 .

1 100  0 .

42  9 .

 0 .

42  9 .

0  100 2 .

1  180

mg = kd 0 , which gives d 0 = mg/k.

The restoring force also leads to simple harmonic motion when the object is attached to a vertical spring, just as it does when the spring is horizontal. When the spring is vertical, however, the weight of the object causes the spring to stretch, and the motion occurs with respect to the equilibrium position of the object on the stretched spring .

Simple Harmonic Motion and the Reference Circle

Simple harmonic motion, like any motion, can be described in terms of displacement, velocity, and acceleration.

DISPLACEMENT

For any object in simple harmonic motion, the time required to complete one cycle is the

period T

Instead of the period, it is more convenient to speak of the

frequency f

of the motion, the frequency being just the number of cycles of the motion per second.

One cycle per second is referred to as one hertz ( Hz ). One thousand cycles per second is called one kilohertz ( kHz ).



is often called the

angular frequency .

VELOCITY velocity is not constant, but varies between maximum and minimum values as time passes

Example 3.

The Maximum Speed of a Loudspeaker Diaphragm

The diaphragm of a loudspeaker moves back and forth in simple harmonic motion to create sound. The frequency of the motion is f = 1.0 kHz and the amplitude is A = 0.20 mm. (a)What is the maximum speed of the diaphragm? (b)Where in the motion does this maximum speed occur?

(a)

(b) The speed of the diaphragm is zero when the diaphragm momentarily comes to rest at either end of its motion: x = +A and x = –A. Its maximum speed occurs midway between these two positions, or at x = 0 m.

Conceptual Example 4.

Moving Lights

Over the entrance to a restaurant is mounted a strip of equally spaced light bulbs, as Figure 10.13a illustrates. Starting at the left end, each bulb turns on in sequence for one-half second. Thus, a lighted bulb appears to move from left to right. Once the apparent motion of a lighted bulb reaches the right side of the sign, the motion reverses. The lighted bulb then appears to move to the left, as part b of the drawing indicates. Thus, the lighted bulb appears to oscillate back and forth. Is the apparent motion simple harmonic motion?

No. Speed is constant.

ACCELERATION

 

a c

cos



a c centripetal acceleration a acceleration of the shadow

a c



 2

(ch8)

Example 5.

The Loudspeaker Revisited—The Maximum Acceleration

A loudspeaker diaphragm is vibrating at a frequency of f = 1.0 kHz, and the amplitude of the motion is A = 0.20 mm. (a)What is the maximum acceleration of the diaphragm, and (b)where does this maximum acceleration occur?

(a)

(b) the maximum acceleration occurs at x = +A and x = –A

FREQUENCY OF VIBRATION

F = ma





cos 

 

 



2

cos 

  

k m

 

in rad

 22

Example 6.

A Body Mass Measurement Device

Astronauts who spend long periods of time in orbit periodically measure their body masses as part of their health-maintenance programs. On earth, it is simple to measure body weight W with a scale and convert it to mass m using the acceleration due to gravity, since W = mg. However, this procedure does not work in orbit, because both the scale and the astronaut are in free-fall and cannot press against each other. Instead, astronauts use a body mass measurement device. This device consists of a spring-mounted chair in which the astronaut sits. The chair is then started oscillating in simple harmonic motion. The period of the motion is measured electronically and is automatically converted into a value of the astronaut’s mass, after the mass of the chair is taken into account. The spring used in one such device has a spring constant of 606 N/m, and the mass of the chair is 12.0 kg. The measured oscillation period is 2.41 s. Find the mass of the astronaut.

  2 



k m

Check Your Understanding 2

The drawing shows plots of the displacement x versus the time t for three objects undergoing simple harmonic motion. Which object, I, II, or III, has the greatest maximum velocity?

Energy and Simple Harmonic Motion

A spring also has potential energy when the spring is stretched or compressed, which we refer to as

elastic potential energy

. Because of elastic potential energy, a stretched or compressed spring can do work on an object that is attached to the spring.

DEFINITION OF ELASTIC POTENTIAL ENERGY The elastic potential energy PE elastic is the energy that a spring has by virtue of being stretched or compressed. For an ideal spring that has a spring constant k and is stretched or compressed by an amount x relative to its unstrained length, the elastic potential energy is

elastic

 1 2

2

SI Unit of Elastic Potential Energy: joule (J)

Example 7.

An Object on a Horizontal Spring

An object of mass m = 0.200 kg that is vibrating on a horizontal frictionless table. The spring has a spring constant k = 545 N/m. It is stretched initially to x 0 = 4.50 cm and then released from rest (see part A of the drawing). Determine the final translational speed v f of the object when the final displacement of the spring is (a) x f = 2.25 cm and (b) x f = 0 cm.

f



0

1 2

 1 2

 2



mgh f

 1 2

 1 2

0 2  1 2

 0 2 

mgh

0  1 2

2 0 1 2

 1 2

 1 2

2 0

v f



k m

(

0 2 

) 32

(a) Since x 0 = 0.0450 m and x f = 0.0225 m, (b) When x 0 = 0.0450 m and x f = 0 m,

Conceptual Example 8.

Changing the Mass of a Simple Harmonic Oscillator

A box of mass m attached to a spring that has a force constant k. The box rests on a horizontal, frictionless surface. The spring is initially stretched to x = A and then released from rest. The box then executes simple harmonic motion that is characterized by a maximum speed v max , an amplitude A, and an angular frequency w.

When the box is passing through the point where the spring is unstrained (x = 0 m), a second box of the same mass m and speed v max is attached to it, as in part b of the drawing. Discuss what happens to (a) the maximum speed, (b) the amplitude, and (c) the angular frequency of the subsequent simple harmonic motion .

 



max



A m

(a) The maximum speed of the two-box system remains the



2

same as that of the one-box system.

(b) The amplitude of the two-box system is greater than that of the one-box system by a factor of

(c) The angular frequency of the two-box system is smaller than that of the one-box system by a factor of

2 35

Example 9.

A Falling Ball on a Vertical Spring

A 0.20-kg ball is attached to a vertical spring. The spring constant of the spring is 28 N/m. The ball, supported initially so that the spring is neither stretched nor compressed, is released from rest. In the absence of air resistance, how far does the ball fall before being brought to a momentary stop by the spring?

= –h

= 2mg/k.

Check Your Understanding 3

A block is attached to the end of a horizontal ideal spring and rests on a frictionless surface. The block is pulled so that the spring stretches relative to its unstrained length. In each of the following three cases, the spring is stretched initially by the same amount, but the block is given different initial speeds. Rank the amplitudes of the resulting simple harmonic motion in decreasing order (largest first). (a) The block is released from rest. (b) The block is given an initial speed v 0 . (c) The block is given an initial speed v 0 /2.

(b), (c), (a)

The Pendulum

simple pendulum

consists of a particle of mass m, attached to a frictionless pivot P by a cable of length L and negligible mass.

Example 10.

Keeping Time

Determine the length of a simple pendulum that will swing back and forth in simple harmonic motion with a period of 1.00 s.

 1 2 

Example 11.

Pendulum Motion and Walking

When we walk, our legs alternately swing forward about the hip joint as a pivot. In this motion the leg is acting approximately as a physical pendulum. Treating the leg as a uniform rod of length D = 0.80 m, find the time it takes for the leg to swing forward.

The desired time is one-half of the period or 0.75 s.

Damped Harmonic Motion

In the presence of energy dissipation, the amplitude of oscillation decreases as time passes, and the motion is no longer simple harmonic motion. Instead, it is referred to as

damped harmonic motion

, the decrease in amplitude being called “damping.”

The smallest degree of damping that completely eliminates the oscillations is termed “critical damping,” and the motion is said to be

critically damped

. When the damping exceeds the critical value, the motion is said to be

overdamped

. In contrast, when the damping is less than the critical level, the motion is said to be

underdamped

(curves 2 and 3).

Conceptual Question 6

REASONING AND SOLUTION A block is attached to a horizontal spring and slides back and forth in simple harmonic motion on a frictionless horizontal surface. A second identical block is suddenly attached to the first block when the first block is at one extreme end of the oscillation cycle. a. Since the attachment is made at one extreme end of the oscillation cycle, where the velocity is zero, the extreme end of the oscillation cycle will remain at the same point; in other words, the amplitude remains the same.

b. The angular frequency of an object of mass m in simple harmonic motion at the end of a spring of force constant k

 

k

/

m

is doubled while the force constant k remains the same, c. The maximum speed of oscillation is given by Equation

v

max 

A



same and the angular frequency, w, decreases by a factor

Conceptual Question 11

REASONING AND SOLUTION From Equations 10.5 and 10.11, we can deduce that the period of the simple harmonic motion of



the end of the ideal spring and k is the spring constant. We can deduce from Equations 10.5 and 10.16 that, for small angles, the is the length of the pendulum. In principle, the motion of a simple pendulum and an object on an ideal spring can both be used to provide the period of a clock. However, it is clear from the expressions for the period given above that the period of the mass-spring system depends only on the mass and the spring constant, while the period of the pendulum depends on the acceleration due to gravity. Therefore, a pendulum clock is likely to become more inaccurate when it is carried to the top of a high mountain where the value of g will be smaller than it is at sea level.

Conceptual Question 12

REASONING AND SOLUTION We can deduce from Equations 10.5 and 10.16 that, for small angles, the period, T,

2 

length of the pendulum. This can be solved for the acceleration due to gravity to yield:

 4  2

If you were held prisoner in a room and had only a watch and a pair of shoes with shoelaces of known length, you could determine whether this room is on earth or on the moon in the following way: You could use one of the shoelaces and one of the shoes to make a pendulum. You could then set the pendulum into oscillation and use the watch to measure the period of the pendulum. The acceleration due to gravity could then be calculated from the expression above. If the value is close to 9.80 m/s 2 , then it can be concluded that the room is on earth. If the value is close to 1.6 m/s 2 , then it can be concluded that the room is on the moon.

Problem 8

REASONING AND SOLUTION The figure at the right shows the original situation before the spring is cut. The weight, W, of the object stretches the string by an amount x. Applying F = kx to this situation gives W = kx (1)

W kx 49

The figure at the right shows the situation after the spring is cut into two segments of equal length.

k'x'

Let k' represent the spring constant of each half of the spring after it is cut. Now the weight, W, of the object stretches each segment by an amount x'.

W

Applying F = kx to this situation gives W = k'x' + k'x' = 2k'x' (2)

k'x'

Combining Equations (1) and (2) yields

kx = 2k'x'

From Conceptual Example 2, we know that k' = 2k so that

kx = 2(2k)x'

Solving for x' gives

'



4



0.160 m 4



0.040 m

Problem 16

REASONING AND SOLUTION From Conceptual Example 2, we know that when the spring is cut in half, the spring constant for each half is twice as large as that of the original spring. In this case, the spring is cut into four shorter springs. Thus, each of the four shorter springs with 25 coils has a spring constant of

4



420 N/m



1680 N/m

The angular frequency of simple harmonic motion is given by Equation 10.11:

 

k m



1680 N/m



46 kg 6.0 rad/s

Problem 17

REASONING AND SOLUTION

a. Since the object oscillates between , the amplitude of the motion is 0.08m

b. From the graph, the period is T=4.0 s . Therefore, according to Equation 10.4,

 

2





2



4.0 s



1.6 rad/s

c. Equation 10.11 relates the angular frequency to the spring

 

k

/

m

  2



(1.6 rad/s)

(0.80 kg)



2.0 N/m

d. At t=1.0 s, the graph shows that the spring has its maximum displacement. At this location, the object is momentarily at rest, so that its speed is v=0 m/s e. The acceleration of the object at t=1.0 s is a maximum, and its magnitude is

max 

 2 

(0.080 m)(1.6 rad/s)

= 0.20 m/s

2 54

Problem 21

REASONING The frequency of vibration of the spring is related to the added mass m by Equations 10.6 and 10.11



1 2



k m

The spring constant can be determined from Equation 10.1

SOLUTION Since the spring stretches by 0.018 m when a 2.8-kg object is suspended from its end, the spring constant is, according to Equation 10.1,



1 2



k m f

2 

1 4

 2

k m m



4

 2

k



F

Applied

x



mg x

 2

(2.8 kg)(9.80 m/s ) 0.018 m

  3

Solving Equation (1) for m, we find that the mass required to make the spring vibrate at 3.0 Hz is



4



2



1.52



10 3 N/m 4



2 (3.0 Hz) 2



4.3 kg

Problem 25

0.392m

2.0kg

0. 2m point of release



elastic

0. 2m 0. 2m

Problem 25

REASONING AND SOLUTION If we neglect air resistance, only the conservative forces of the spring and gravity act on the ball. Therefore, the principle of conservation of mechanical energy applies When the 2.00 kg object is hung on the end of the vertical spring, it stretches the spring by an amount x, where



F k



mg k



(2.00 kg)(9.80 m/s

) 50.0 N/m



0.392 m

This position represents the equilibrium position of the system with the 2.00-kg object suspended from the spring. The object is then pulled down another 0.200 m and released from rest (v 0 =0 m/s).

At this point the spring is stretched by an amount h = 0 m: The kinetic energy, the gravitational potential energy, and the elastic potential energy at the point of release are:

KE

 1 2

0 2  1 2

(0 m/s)

2 

0 J PE

gravity 

mgh



(0 m)



0 J PE

elastic  1 2

kx

0 2  1 2

(50.0 N/m)(0.592 m)

2 

8.76 J

The total mechanical energy E 0 at the point of release is the sum of the three energies above:

0  8.76 J 59

above the release point, the spring is stretched by an amount The gravitational and elastic potential energies are:

PE gravity 

mgh

 (2.00 kg)(9.80 m/s 2 )(0.200 m)  3.92 J

PE

elastic  1 2

kx

2  1 2

(50.0 N/m)(0.392 m)

2 

3.84 J

Since

KE



PE

gravity 

PE

elastic 

KE  E – PE gravity – PE elastic  8.76 J – 3.92 J – 3.84 J = 1.00 J 60

above the release point, the spring is stretched by an amount of . At this point, the total

8.76 J

elastic potential energies are:

PE

gravity 

mgh



(2.00 kg)(9.80 m/s

)(0.400 m)



7.84 J PE

elastic  1 2

2  1 2

(50.0 N/m)(0.192 m)

2 

0.92 J

The kinetic energy is





PE gravity



PE elastic

 8 .

 7 .

 0 .

 0

The results are summarized in the table below

0.000 m 0.200 m 0.400 m KE 0.00 J 1.00 J 0.00 J PE grav 0.00 J 3.92 J 7.84 J PE elastic 8.76 J 3.84 J 0.92 J

8.76 J 8.76 J 8.76 J

Problem 32

f = 3.0 H Z

A = 5.08

* 10 -2 m

m /2

Max speed at halfway of the amplitude.

m m /2

REASONING AND SOLUTION

a. Now look at conservation of energy before and after the split Before split (1/2) mv max 2 = (1/2) kA 2 Solving for the amplitude A gives

A  v max m k

After split If new amplitude is A’ (1/2) (m/2)v' 2 = (1/2) (m/2)(v max ) 2 = (1/2) kA' 2

Solving for the amplitude A' gives

A

 

v

max

m 2k

Therefore, we find that A' = A/ = (5.08 * 10 –2 m)/ = 3.59

* 10 -2 m

 

k m

2





k m f

 1 2 

Similarly, for the frequency, we can show that

k m

b. If the block splits at one of the extreme positions, the amplitude of the SHM would not change, so it would remain as 5.08

* 10 -2 m The frequency would be

Problem 38

t = 0.25 s

Object is resting on the spring.

F = kx = mg



g k m



0 .

015 sec

ond

REASONING AND SOLUTION

Using f = 1/T = 1/(0.250 s) = 4.00 Hz and also

f = 1 2



k m



2 f 2 = 632 N/(kg



m) With the object resting on the spring, F = kx = mg so that,

x = g k m = 0.0155 m.

When the mass leaves the spring, potential energy of the spring has been converted to gravitational energy, i.e., if h is the height, it can reach (1/2) kx' 2 = mgh Where x' = 0.0500 m + 0.0155 m = 0.0655 m Solving for h we get

h

  

k m

   

x' 2g

2  

632 N/(kg



m)

 

(0.0655 m)

2 

2(9.80 m/s

)



0.138 m

Problem 41

REASONING AND SOLUTION Recall that the relationship between frequency f and period T is . Then, according to Equations 10.6 and 10.16, the period of the simple pendulum is given by

T



2



L g

where L is the length of the pendulum. Solving for g and noting that the period is T = (280 s)/100 = 2.8 s, we obtain



4



2

L T

2



4



2 (1.2 m) (2.8 s) 2



6.0 m/s 2

Problem 68

REASONING The force F that the spring exerts on the block just before it is released is equal to –kx, according to Equation 10.2. Here k is the spring constant and x is the displacement of the spring from its equilibrium position. Once the block has been released, it oscillates back and forth with an angular



the mass of the block. The maximum speed that the block attains during the oscillatory motion is v max = A



(Equation 10.8). The magnitude of the maximum acceleration that the block attains is a max



2 (Equation 10.10).

SOLUTION

a. The force F exerted on the block by the spring is

 

     

9.84 N

motion is

 

k m



82.0 N /m



0.750 kg 10.5 rad /s

c. The maximum speed v max is the product of the amplitude and the angular frequency:

max 

    

1.26 m /s

d. The magnitude a max of the maximum acceleration is

max 

 2    2 

13.2 m /s

2 73