chapter 2 motion along a straight line position, displacement, … · 2017. 2. 14. · title:...

We can pull the mass to the right and then release it to begin its motion:

Or we could push it to the left and release it:

Both motions would have the same values for T and f.

However, the resulting motion will have a phase difference of half a cycle.

Start stretched

Start compressed

The two motions are half a cycle out of phase.

x

x

Phase difference

𝑜𝑟 𝜋

PRACTICE: Two identical mass-spring systems are started in two different ways. What is their phase difference?

SOLUTION:

The phase difference is one-quarter of a cycle 𝜋

2.

Start stretched and then release

Start unstretched with a push left

x

x

Phase difference

PRACTICE: Two identical mass-spring systems are started in two different ways. What is their phase difference?

SOLUTION:

The phase difference is three-quarters of a cycle 3𝜋

2

Start stretched and then release

Start unstretched with a push right

x

x

Phase difference

x

Conditions for simple harmonic motion

EXAMPLE: A spring having a spring constant of 125 N m-1 is attached to a 5.0-kg mass, stretched +4.0 m as shown, and then released from rest.

(a) Using Hooke’s law, show that the acceleration a of a mass-spring system is related to the spring’s displacement x by the proportion a -x.

F = -kx & F = ma

ma = -kx

a = -(k / m) x.

∴ a -x.

(b) Taylor your equation to this example, and find the acceleration of the mass when x = -2.0 m.

a = -(k / m) x

a = -25x

a = -25(-2.0) = +50. ms-2.

(c) What is the displacement of the mass when the acceleration is

-42 ms-2?

a = -25x

x = -a/25

x = +1.7 m

Sketching and interpreting graphs of SHM examples

EXAMPLE: The displacement x vs. t for a 2.5-kg mass on a spring having spring constant k = 4.0 Nm-1 is shown in the sinusoidal graph.

SOLUTION: The period is the time for a complete cycle.

(a) Find the period and frequencyof the motion.

(b) Find the amplitude of the motion.

From the graph it is xMAX = 2.0 mm = 2.010-3 m.

SOLUTION: The amplitude is the maximum displacement.

From the graph it is T = 6.0 ms = 6.010-3 s.

Then f = 1 / T = 1 / 0.006 = 170 Hz.

(c) Sketch the graph of x vs. t for the situation where the amplitude is cut in half.

SOLUTION: For SHM, the period is independent of the amplitude.

(c) The blue graph shows an equivalent system in SHM. What is the phase difference between the red and blue?

SOLUTION:

We see that it is T / 6 (= 360/6 = 60 = 2/6 rad).

Sketching and interpreting graphs of SHM examples

EXAMPLE: The kinetic energy vs. displacement fora system undergoing SHM is shown. The system consists of a 0.125-kg mass on a spring.

(a) Determine the maximum velocity of the mass.

SOLUTION: When the Ek is max, the velocity is also max.

𝐸𝑘 𝑚𝑎𝑥 =1

2𝑚 𝑣𝑚𝑎𝑥

2 → 4.0 =1

20.125 𝑣𝑚𝑎𝑥

2

vMAX = 8.0 ms-1.

(b) Sketch EP and determine the total energy of the system.

SOLUTION: Since EK + EP = ET = CONST, EP = 0 when EK = 𝐸𝑘 𝑚𝑎𝑥, it must be that ET = 𝐸𝑘 𝑚𝑎𝑥 = 4.0 J

Thus the EP graph will be the “inverted” EK graph

ET

EP

EK

http://en.wikipedia.org/wiki/File:Simple_harmonic_oscillator.gifhttp://en.wikipedia.org/wiki/File:Simple_harmonic_oscillator.gif

(c) Determine the spring constant k of the spring.

SOLUTION: EP = (1/2)kx2. EK = 0 at x = xMAX = 2.0 cm. EK + EP = ET = CONST

ET = 0 + (1/ 2)kxMAX2 4.0 = (1/ 2)k 0.0202

k = 20000 Nm-1.

(d) Determine the acceleration of the mass at x = 1.0 cm.

SOLUTION: From Hooke’s law, F = -kx

F = -20000(0.01) = -200 N. From F = ma

-200 = 0.125aa = -1600 ms-2.

Sketching and interpreting graphs of SHM examplesEXAMPLE: A 4.0-kg mass is placed on a spring’s end

and displaced 2.0 m to the right.The spring force F vs. its displacement x from equilibrium is shown in the graph.

(a) How do you know that the mass is undergoing SHM?

SOLUTION: In SHM, a -x. Since F = ma, then F -x also. The graph shows that F -x. Thus we have SHM.

(b) Find the spring constant of the spring.

SOLUTION: Hooke’s law: F = -kx.

Pick any F and any x. Use k = -F / x.

F = -5.0 Nx = 1.0 m

k = -(-5.0 N) / 1.0 m = 5.0 Nm-1.

(c) Find the total energy of the system.

SOLUTION: ET = (1/2)kxMAX2. Then ET = (1/2)kxMAX

2 = (1/2) 5.0 2.02 = 10. J.

(d) Find the maximum speed of the mass.

SOLUTION: ET = (1/2)mvMAX2.

10. = (1/2) 4.0 vMAX2

vMAX = 2.2 ms-1

(e) Find the speed of the masswhen its displacement is 1.0 m.

SOLUTION: ET = (1/2)mv2 + (1/2)kx 2.

Then 10. = (1/2)(4)v 2 + (1/2)(5)12

v = 1.9 ms-1.

PRACTICE: A spring is moved in SHM by the hand as shown. The hand moves through 1.0 complete cycle in 0.25 s. A metric ruler is placed beside the waveform.

(a) What is the wavelength?

= 4.7 cm = 0.047 m.

(b) What is the period?

T = 0.25 s.

(c) What is the wave speed?

v = / T = 0.047 / 0.25 = 0.19 m s-1.

1234567891011121314 CM

Solving wave speed and wavelength problems

Either graph gives the correct amplitude.

Sketching and interpreting displacement vs time graphs

EXAMPLE:

Graph 1 shows the variation with time t of the displacement d of a traveling wave.

Graph 2 shows the variation with distance x along the same wave of its displacement d.

(a) Use the graphs to determine the amplitude of the wave motion.

Amplitude (maximum displacement) is 0.0040 m.

(b) Use the graphs to determine the wavelength.

Wavelength is measured in meters and is the length of a complete wave. = 2.40 cm = 0.0240 m.

Graph 2 must be used since its horizontal axis is in cm (not seconds

as in Graph 1).

(c) Use the graphs to determine the period.

Period is measured in seconds and is the time for one complete wave. T = 0.30 s.

(d) Use the graphs to find the frequency.

f = 1 / T = 1 / 0.30 = 3.3 Hz. [3.333 Hz]

Graph 1 must be used since its horizontal axis is in s (not cm as in

Graph 2).

(e) Use the graphs to find the wave speed.

v = / T = 0.024 / 0.30 = 0.080 m s-1.

Graph 2 must be used for since its horizontal axis is

in cm.


PRACTICE: Graph 1 shows the variation with time t of the displacement y of a traveling wave.

Graph 2 shows the variation with distance x along the same wave of its displacement.

(a) Use the graphs to determine the amplitudeand wavelength of the wave motion.

Amplitude (maximum displacement) is y = 0.0020 m. Wavelength is y = 0.30 cm = .0030 m.

(b) Use the graphs to determine the periodand the frequency.

(c) Use the graphs to determine the wave speed.

Period (cycle time) is 0.25 ms = 0.00025 s. Frequency is f = 1 / T = 1 / 0.00025 = 4000 Hz.

Wave speed is a calculation. v = / T = 0.0030 / 0.00025 = 12 m s-1.

Graph 1 must be used for T since its horizontal axis is

in ms.

Sketching and interpreting displacement vs time graphsEXAMPLE:Graph 1 shows the variation with time t of the displacement x of a single particle in the medium carrying a longitudinal wave in the +x direction.

(a) Use the graph to determine the period and the frequency of the particle’s SHM.

The period is the time for one cycle. T = 0.20 s. f = 1 / T = 1 / 0.20 = 5.0 Hz.

Graph 2 shows the variation of the displacement x with distance d from the beginning of the wave at a particular instant in time.

(b) Use the graph to determine the wavelength and wave velocity of the longitudinal wave motion.

= 16.0 cm = 0.160 m.

v = / T = 0.160 / 0.20 = 0.80 m s-1.


(c) The equilibrium positions of 6 particles in the medium are shown below. Using ’s, indicate the actual position of each particle at the instant shown above.

(d) In the diagram label the center of a compression with a C and the center of a rarefaction with an R.

C R

A sound wave produced by a clock chime is

heard 515 m away 1.5 s later.

(a)What is the speed of sound in the air there?

(b) The sound wave has a frequency of 436 Hz.

What is the period of the wave?

(c) What is the wave's wavelength?

(a) v = d/t = 515/1.5 = 343 m/s

(b) f = 1/T = 1/436 = 2.29x10-3 s

(c) v = f → = v / f = 0.87 m

http://images.google.com/imgres?imgurl=http://i165.photobucket.com/albums/u71/timrel/Skeleton-Wall-Clock-0901X-1.jpg&imgrefurl=http://www.clockshopuk.co.uk/ouronlineshopclockshopuk/prod_268490-Heinsberg-Clock.html&usg=__LMNDyTq1R1sBbFt1nlRlVq3TyAQ=&h=900&w=420&sz=84&hl=en&start=69&tbnid=NCBtCNkROxK6KM:&tbnh=146&tbnw=68&prev=/images?q=clock+chime&start=60&ndsp=20&hl=en&sa=Nhttp://images.google.com/imgres?imgurl=http://i165.photobucket.com/albums/u71/timrel/Skeleton-Wall-Clock-0901X-1.jpg&imgrefurl=http://www.clockshopuk.co.uk/ouronlineshopclockshopuk/prod_268490-Heinsberg-Clock.html&usg=__LMNDyTq1R1sBbFt1nlRlVq3TyAQ=&h=900&w=420&sz=84&hl=en&start=69&tbnid=NCBtCNkROxK6KM:&tbnh=146&tbnw=68&prev=/images?q=clock+chime&start=60&ndsp=20&hl=en&sa=N

A hiker shouts toward a vertical cliff 465 m away. The echo is heard

2.75 s later. (a) What is the speed of sound in air there? (b) The

wavelength of the sound is 0.75 m. What is the frequency of the

wave? (c) What is its period?

(b) v = f → f = v/ = 338/0.75 = 451 Hz

(a) v = distance/time = 2d/t = 2·465/2.75 = 338 m/s

(c) T = 1/f = 2.22x10-3 s

If you wanted to increase the wavelength of waves in a rope

should you shake it at a higher or lower frequency?

v = f v depends only on the medium. Therefore for given

medium it is constant. So if wavelength increases the

frequency decreases. You should shake it at lower

frequencies. CHECK IT, PLEASE

A stone is thrown onto a still water surface and creates a wave.

A small floating cork 1.0 m away from the impact point has the

following displacement—time graph (time is measured from the

instant the stone hits the water):

Find (a) amplitude (b) the speed of the wave (c) the freq. (d) wavelength

(a) A = 2 cm

(b) v = d/t = 1/1.5 = 0.67 m/s

(c) f = 1/T = 1/0.3 = 3.33 Hz

(d) λ = v/f = 0.666/3.333 = 0.2 m

http://upload.wikimedia.org/wikipedia/commons/4/43/2006-01-14_Surface_waves.jpghttp://upload.wikimedia.org/wikipedia/commons/4/43/2006-01-14_Surface_waves.jpghttp://upload.wikimedia.org/wikipedia/commons/4/43/2006-01-14_Surface_waves.jpghttp://upload.wikimedia.org/wikipedia/commons/4/43/2006-01-14_Surface_waves.jpghttp://upload.wikimedia.org/wikipedia/commons/4/43/2006-01-14_Surface_waves.jpghttp://upload.wikimedia.org/wikipedia/commons/4/43/2006-01-14_Surface_waves.jpg

SOLUTION: = c / f = 3.00108 / 1.6710 15 = 1.8010 -7 m.

(a) What is its frequency, and what part of the spectrum is it from?

SOLUTION: T = 6.0010 -16 s.

f = 1 / T = 1 / 6.00 10 -16 s = 1.6710 15 Hz.

This is from the ultraviolet part of the spectrum.

(b) What is the wavelength of this light wave?

PRACTICE: The graph shows one complete oscillation of a particular frequency of light.

(c) Determine whether or not this light is in the visible spectrum.

SOLUTION: The visible spectrum is from about 400 nm to 700 nm.

= 1.8010 -7 m = 18010 -9 m = 180 nm. NO! UV.

700 600 500 400

Wavelength / nm

Solving problems involving amplitude and intensity

EXAMPLE: A 200. watt speaker projects sound in a spherical wave. Find the intensity of the sound at a distance of 1.0 m and 2.0 m from the speaker.

Note that whatever power is in the wavefront is spread out over a larger area as it expands.

The area of a sphere of radius x is A = 4x2.

For x = 1 m: I = P / (4x2) = 200 / (41.02) = 16 W m-2.

For x = 2 m: I = P / (4x2) = 200 / (42.02) = 4.0 W m-2.

Doubling your distance reduces the intensity by 75%!

Solving problems involving amplitude and intensity

EX: At a distance of 18.5 m from a sound source the intensity is 2.0010 -1 W m-2.

(a) Find its intensity at a distance of 26.5 m.

Then I1 x1-2 and I2 x2

-2 so that

I2 / I1 = x2-2 / x1-2 = x12 / x22 = (x1 / x2)2.

Thus

I2 = I1 (x1 / x2)2

= 2.0010 -1 (18.5 / 26.5)2 = 0.0975 W m -2.

SOLUTION: We can just use I x -2 and dispense with finding the actual power as an intermediate step.

SOLUTION: We can use I A 2 and I1 and I2.

Then I1 A12 and I2 A2

2 so that

I1 / I2 = A12/ A2

2.

Then

A1 / A2 = 𝐼1 / 𝐼2

= 2.0010 −1 / 0.0975= 1.43.

A1 = 1.43 A2.

(b) Compare the amplitudes of the sound at 18.5 m and 26.5 m.

PRACTICE: Two pulses are shown in a string approaching each other at 1 m s-1.

Sketch diagrams to show each of the following:

(a) The shape of the string at exactly t = 0.5 s later.

(b) The shape of the string at exactly t = 1.0 s later.

1 m1 m

1 m

Since the pulses are 1 m apart and approaching each other at 1 ms-1 they will overlap at this instant, temporarily canceling out.

Since the pulses are 1 m apart and approaching each other at 1 m s-1, at this time they will each have moved ½ m and their leading edges will just meet.

Superposition

EXAMPLE:

Two waves P and Q reach the same point at the same time, as shown in the graph.

The amplitude of the resulting wave is

A. 0.0 mm. B. 1.0 mm. C. 1.4 mm. D. 2.0 mm.

In the orange regions P and Q do not cancel.

In the purple regions P and Q partially cancel.

Focus on orange. Try various combos…

Blue (0.7 mm) plus green (0.7 mm) = 1.4 mm.

Note that the crests to not coincide, so NOT 2.0.

Superposition

Observe the

light intensity

through two

polarizing filters

as the top one is

rotated through

180º.

Which filter is

the analyzer?

Which filter is

the polarizer?

RED FILTER

BLACK FILTER

Solving problems involving Malus’s law

PRACTICE: The preferred directions of two sheets of Polaroid are initially parallel.(a) Calculate the angle through which one sheet needs to be turned in order to reduce the amplitude of the observed E-field to half its original value.(b) Calculate the effect this rotation has on the intensity.(c) Calculate the rotation angle needed to halve the intensity from its original value.

SOLUTION:

(a) Solve cos = 1 / 2 = 60º.

(b) I = I0 cos2 = I0(1/2)

2 = I0 / 4.

(c) I0 / 2 = I0 cos2 cos2 = 1 / 2. Thus

cos = (1 / 2)1/2 = 450 .


In general, light sources produce waves having their E-fields oriented in many random directions.

Polarized light is light whose waves have their E-fields oriented in one direction.

I = I0 cos2 ⇒ I = I0 cos

2 600 ⇒ I = 0.25 I0 .

I0 cos2 0º = I0

I0 cos2 60º = 0.25I0

I0 cos2 90º = 0

I0 cos2 120º = 0.25I0

I0 cos2 180º = I0

In general, light sources produce waves having their E-fields oriented in many random directions.

Polarizing sunglasses only allow waves in one direction through, thereby reducing the intensity of the light entering the eye.

Reflecting surfaces also polarize light to a certain extent, thereby rendering polarizing sunglasses even more effective.


I = I0 cos2 ⇒ I = I0 (1/2)

2 = I0 / 4

Diffraction allows waves to turn corners.

Wave behaviorAll waves diffract – not just sound.

Huygens says the wavelets will be spherical.

Wave behavior

Wave behavior

PRACTICE: What wave behavior is being demonstrated

here?

Refraction.

T or F: The period changes

in the different media.

T or F: The frequency

changes in the different media.

T or F: The wavelength

changes in the different media.

T or F: The wave speed changes in the different media.

T or F: The sound wave is traveling fastest in the warm

air.

A sound pulse entering

and leaving a pocket

of cold air (blue).

Wave behavior

PRACTICE: What wave behavior is being demonstrated

here?

Reflection. This is an echo from a wall.

T or F: The frequency of the reflected wave is different

from the frequency of the incident wave?

Fill in the blanks: The angle of

___________ equals the angle

of ___________.

Fill in the blank: In this example, both angles equal ___°.

Remember that angles are measured

wrt the normal.

incidence

reflection

0

Wave behavior

PRACTICE: What wave behavior is being demonstrated here?

Total internal reflection.

This is the same principle behind fiber optics. Light is used instead of sound.

Light travels faster than electrical signals so fiber optics are used in high speed communications networks such as token rings whenever possible.

Wave behavior

PRACTICE: Complete with sketch and formula.

Remember that there are always nodes on each end of

a string.

Add a new well-spaced node each time.

Decide the relationship between and L.

We see that = (2/3)L.

Since v = f we see that f = v / (2/3)L = 3v / 2L.

f2 = v / L

f1 = v / 2L

f3 = 3v / 2L

Sketching and interpreting standing wave patterns

Distinguishing between standing and traveling waves A standing wave consists of two traveling waves carrying energy in opposite directions, so the net energy flow through the wave is zero.

E

E

Solving problems involving standing waves

PRACTICE: A tube is filled with water and a vibrating tuning fork is held above the open end. As the water runs out of the tap at the bottom sound is loudest when the water level is a distance x from the top. The next loudest sound comes when the water level is at a distance y from the top. Which expression for is correct?

A. = x B. = 2x C. = y-x D. = 2(y-x)

v = f and since v and f are constant, so is .

The first possible standing wave is sketched.

The second possible standing wave is sketched.

Notice that y – x is half a wavelength.

Thus the answer is = 2(y - x).

y-x

PRACTICE: This drum head, set to vibrating at different resonant frequencies, has black sand on it, which reveals 2D standing waves.

Does the sand reveal nodes, or does it reveal antinodes?

Why does the edge have to be a node?

Nodes, because there is no displacement to throw the sand off.

The drumhead cannot vibrate at the edge.


Alternate lobes have a 180º phase difference.


Make a sketch. Then use v = f.

antinode antinode

L

/ 2 = L v = f

= 2L f = v /

f = v / (2L)


Reflection provides for two coherent waves traveling in opposite directions.

Superposition is just the adding of the two waves to produce the single stationary wave.


The figure shows the points between successive nodes.

For every point between the two nodes f is the same.

But the amplitudes are all different.

Therefore the energies are also different.


Energy transfer via a vibrating medium without interruption.

The medium itself does not travel with the wave disturbance.

Speed at which the wave disturbance propagates.

Speed at which the wave front travels.

Speed at which the energy is transferred.


Frequency is number of vibrations per unit time.

Distance between successive crests (or troughs).

Distance traveled by the wave in one oscillation of the source.

FYI: IB frowns on you using particular units as in“Frequency is number of vibrations per second.”

FYI: There will be lost points, people!


The waves traveling in opposite directions carry energy at same rate both ways. NO energy transfer.

The amplitude is always the same for any point in a standing wave.


LP / 4 = L

= 4L

L

Q / 2 = L

= 2L

v = f

f = v /

fP = v / (4L)fQ = v / (2L)

v = 4LfPfQ = 4LfP / (2L)

fQ = 2fP


The tuning fork is the driving oscillator (and is at the top).

The top is thus an antinode.

The bottom “wall” of water allows NO oscillation.

The bottom is thus a node.


Sound is a longitudinal wave.

Displacement is small at P, big at Q.


If the lobe at T is going down, so is the lobe at U.


Pattern 1 is 1/2 wavelength.

Pattern 2 is 3/2 wavelength.

Thus f2 = 3f1 so that f1 / f2 = 1/3.


chapter 2 motion along a straight line position, displacement, … · 2017. 2. 14. · title:...

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