Assignment 23

Physics II

06-28-2008

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23:34:24

query doppler shift experiment (experiment was to be read

and viewed only) **** explain why the frequency of the

sound observed when the buzzer moves toward you is greater

than that of the stationary buzzer and why this frequency is

greater than that observed when the buzzer is moving away

from you

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RESPONSE -->

The pulses are coming in at a shorter time interval then

what they are actually being sent at and the frequency is

high, therefore, when the last pulse occurs, it was moving

away making it sound farther away making the frequency be

lower than in the beginning.

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23:34:37

** The 'pulses' emitted by an approaching source in a

certain time interval are all received in a shorter time

interval, since the last 'pulse' is emitted closer to the

source than the first and therefore arrives sooner than if

the source was still. So the frequency is higher.

If the source is moving away then the last 'pulse' is

emitted further from the source than if the source was

still, hence arrives later, so the pulses are spread out

over a longer time interval and the frequency is lower.

GOOD EXPLANATION FROM STUDENT:

Well, for the purposes of this explanation, I am going to

explain the movement of the buzzer in one dimention which

will be towards and away. The buzzer is actually moving in

a circle which means it exists in three dimentions but is

moving in two dimentions with relation to the listener.

However, using trigonometry we can determine that at almost

all times the buzzer is moving either towards or away from

the listener so I will explain this in terms of one

dimention.

}When a buzzer is 'buzzing' it is emitting sound waves at a

certain frequency. This frequency appears to change when

the buzzer moves toward or away from the listener but the

actual frequency never changes from the original frequency.

By frequency we mean that a certain number of sound waves

are emitted in a given time interval (usually x number of

cycles in a second). So since each of the waves travel at

the same velocity they will arrive at a certain vantage

point at the same frequency that they are emitted. So If

a 'listener' were at this given vantage point 'listening',

then the listener would percieve the frequency to be what it

actually is. Now, if the buzzer were moving toward the

listener then the actual frequency being emitted by the

buzzer would remain the same. However, the frequency

percieved by the listener would be higher than the actual

frequency. This is because, at rest or when the buzzer is

not moving, all of the waves that are emitted are traveling

at the same velocity and are emitted from the same location

so they all travel the same distance. But, when the buzzer

is moving toward the listener, the waves are still emitted

at the same frequency, and the waves still travel at the

same velocity, but the buzzer is moving toward the listener,

so when a wave is emitted the buzzer closes the distance

between it and the listener a little bit and there fore the

next wave emitted travels less distance than the previous

wave. So the end result is that each wave takes less time

to reach the listener than the previously emitted wave.

This means that more waves will reach the listener in a

given time interval than when the buzzer was at rest even

though the waves are still being emitted at the same rate.

This is why the frequency is percieved to be higher when the

buzzer is moving toward the listener.

By the same token, if the same buzzer were moving away from

the listener then the actual frequency of the waves emitted

from the buzzer would be the same as if it were at rest, but

the frequency percieved by the listener will be lower than

the actual frequency. This is because, again at rest the

actual frequency will be the percieved frequency. But when

the buzzer is moving away from the listener, the actual

frequency stays the same, the velocity of the waves stays

the same, but because the buzzer moves away from the

listener a little bit more each time it emits a wave, the

distance that each wave must travel is a little bit more

than the previously emitted wave. So therefore, less waves

will pass by the listener in a given time interval than if

the buzzer were not moving. This will result in a lower

percieved frequency than the actual frequency. **

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RESPONSE -->

I understand that pulses' emitted by an approaching source

in a certain time interval are all received in a shorter

time interval, since the last 'pulse' is emitted closer to

the source than the first and therefore arrives sooner than

if the source was still. So the frequency is higher.

If the source is moving away then the last 'pulse' is

emitted further from the source than if the source was

still, hence arrives later, so the pulses are spread out

over a longer time interval and the frequency is lower.

GOOD EXPLANATION FROM STUDENT:

Well, for the purposes of this explanation, I am going to

explain the movement of the buzzer in one dimention which

will be towards and away. The buzzer is actually moving in

a circle which means it exists in three dimentions but is

moving in two dimentions with relation to the listener.

However, using trigonometry we can determine that at almost

all times the buzzer is moving either towards or away from

the listener so I will explain this in terms of one

dimention.

}When a buzzer is 'buzzing' it is emitting sound waves at a

certain frequency. This frequency appears to change when

the buzzer moves toward or away from the listener but the

actual frequency never changes from the original frequency.

By frequency we mean that a certain number of sound waves

are emitted in a given time interval (usually x number of

cycles in a second). So since each of the waves travel at

the same velocity they will arrive at a certain vantage

point at the same frequency that they are emitted. So If

a 'listener' were at this given vantage point 'listening',

then the listener would percieve the frequency to be what it

actually is. Now, if the buzzer were moving toward the

listener then the actual frequency being emitted by the

buzzer would remain the same. However, the frequency

percieved by the listener would be higher than the actual

frequency. This is because, at rest or when the buzzer is

not moving, all of the waves that are emitted are traveling

at the same velocity and are emitted from the same location

so they all travel the same distance. But, when the buzzer

is moving toward the listener, the waves are still emitted

at the same frequency, and the waves still travel at the

same velocity, but the buzzer is moving toward the listener,

so when a wave is emitted the buzzer closes the distance

between it and the listener a little bit and there fore the

next wave emitted travels less distance than the previous

wave. So the end result is that each wave takes less time

to reach the listener than the previously emitted wave.

This means that more waves will reach the listener in a

given time interval than when the buzzer was at rest even

though the waves are still being emitted at the same rate.

This is why the frequency is percieved to be higher when the

buzzer is moving toward the listener.

By the same token, if the same buzzer were moving away from

the listener then the actual frequency of the waves emitted

from the buzzer would be the same as if it were at rest, but

the frequency percieved by the listener will be lower than

the actual frequency. This is because, again at rest the

actual frequency will be the percieved frequency. But when

the buzzer is moving away from the listener, the actual

frequency stays the same, the velocity of the waves stays

the same, but because the buzzer moves away from the

listener a little bit more each time it emits a wave, the

distance that each wave must travel is a little bit more

than the previously emitted wave. So therefore, less waves

will pass by the listener in a given time interval than if

the buzzer were not moving. This will result in a lower

percieved frequency than the actual frequency. **

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23:34:40

query General College Physics and Principles of Physics:

what is a decibel?

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RESPONSE -->

dB=10log(I/I0)

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23:34:51

** dB = 10 log( I / I0 ), where I is the intensity of the

sound in units of power per unit area and I0 is the 'hearing

threshold' intensity.

MORE EXTENSIVE EXPLANATION FROM STUDENT:

Sound is possible because we exist in a medium of air. When

a sound is emitted, a concussive force displaces the air

around it and some amount energy is transferred into kinetic

energy as air particles are smacked away from the force.

These particles are now moving away from the initial force

and collide into other air particles and send them moving

and ultimately through a series of collisions the kinetic

energy is traveling out in all directions and the air

particles are what is carrying it. The behavior of this

kinetic energy is to travel in waves. These waves each

carry some amount of kinetic energy and the amount of energy

that they carry is the intensity of the waves. Intensities

of waves are given as a unit of power which is watts per

square meter. Or since the waves travel in all directions

they move in three dimentions and this unit measures how

many watts of energy hits a square meter of the surface

which is measuring the intensity. But we as humans don't

percieve the intensities of sound as they really are. For

example, a human ear would percieve sound B to be twice as

loud as sound A when sound B is actually 10 times as loud

as sound A. Or a sound that is ...

1.0 * 10^-10 W/m^2 is actually 10 times louder than a sound

that is

1.0 * 10^-11 W/m^2 but the human ear would percieve it to

only be twice as loud.

The decibel is a unit of intensity for sound that measures

the intensity in terms of how it is percieved to the human

ear. Alexander Graham Bell invented the decibel. Bell

originally invented the bel which is also a unit of

intensity for waves. The decibel is one tenth of a bel and

is more commonly used. The formula for determing the

intensity in decibels is ...

Intensity in decibles = the logarithm to the base 10 of the

sound's intensity/ I base 0

I base 0 is the intensity of some reference level and is

usually taken as the minimum intensity audible to an average

person which is also called the 'threshold of hearing'.

Since the threshold of hearing is in the denominator, if a

sound is this low or lower the resulting intensity will be 0

decibles or inaudible. **

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RESPONSE -->

I understand that dB = 10 log( I / I0 ), where I is the

intensity of the sound in units of power per unit area and

I0 is the 'hearing threshold' intensity.

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23:34:55

gen phy what is the difference between the node-antinode

structure of the harmonics a standing wave in a string and

in an organ pipe closed at one end

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RESPONSE -->

I know that in a string, there is nodes at both ends, but I

am unsure in an organ pipe closed at one end

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23:35:06

** in a string there are nodes at both ends so the harmonics

are described the the configurations NAN, NANAN, NANANAN,

etc.. In a pipe closed at one end there is a node at one

end and an antinode at the other so the possible

configurations are NA, NANA, NANANA, etc..

displacement nodes are at both ends of the string, so the

structure is N &&& N, where &&& is any sequence of nodes and

antinodes that results in an alternating sequence.

The possibilities are NAN, NANAN, NANANAN, ... , containing

1, 2, 3, 4, ..., half-wavelengths in the length of the

string. Possible wavelengths are therefore 2 L, 1 L, 2/3

L, ..., where L is string length.

For an open organ pipe the configuration must be N &&& A.

Possibilities include NA, NANA, NANANA, NANANANA, ...,

containing 1, 3, 5, 7, ..., quarter-wavelengths. Possible

wavelengths are therefore 4 L, 4/3 L, 4/5 L, 4/7 L, ... **

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RESPONSE -->

I understand that in a string there are nodes at both ends

so the harmonics are described the the configurations NAN,

NANAN, NANANAN, etc.. In a pipe closed at one end there is

a node at one end and an antinode at the other so the

possible configurations are NA, NANA, NANANA, etc..

displacement nodes are at both ends of the string, so the

structure is N &&& N, where &&& is any sequence of nodes and

antinodes that results in an alternating sequence.

The possibilities are NAN, NANAN, NANANAN, ... , containing

1, 2, 3, 4, ..., half-wavelengths in the length of the

string. Possible wavelengths are therefore 2 L, 1 L, 2/3

L, ..., where L is string length.

For an open organ pipe the configuration must be N &&& A.

Possibilities include NA, NANA, NANANA, NANANANA, ...,

containing 1, 3, 5, 7, ..., quarter-wavelengths. Possible

wavelengths are therefore 4 L, 4/3 L, 4/5 L, 4/7 L, ...

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23:35:09

**** gen phy what are beats?

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RESPONSE -->

beats occur when two sounds are closed in frequency

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23:35:20

** Beats are what happens when the two sounds are close in

frequency. Beats occur when the combined sound gets louder

then quieter then louder etc. with a frequency equal to the

differences of the frequencies of the two sounds. **

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RESPONSE -->

I understand that beats are what happens when the two sounds

are close in frequency. Beats occur when the combined sound

gets louder then quieter then louder etc. with a frequency

equal to the differences of the frequencies of the two sounds

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23:35:25

**** query univ phy 16.54 (20.32 10th edition) steel

rod 1.5 m why hold only at middle to get fund? freq of

fund? freq of 1st overtone and where held? **** why can

the rod be held only at middle to get the fundamental?

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RESPONSE -->

does not apply to me

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23:35:39

STUDENT SOLUTION WITH INSTRUCTOR COMMENTS:

Since the rod is a rigid body it can form the smallest

frequency corresponding to the largest wavelength only when

held at the center. The smallest frequency corresponding to

the largest wavelength is known as the fundamental.

(a) the fingers act as a sort of pivot and the bar performs

a teeter - totter motion on either side of the pivot

**note: the motion is more like the flapping of a bird's

wings, though it obviously doesn't result from the same sort

of muscular action but rather from an elastic response to a

disturbance, and the motion attenuates over time**. With

this manner of motion, the ends have the greatest amount of

amplitude, thus making them the antinodes.

(b)Holding the rod at any point other than the center

changes the wavelngth of the first harmonic causing one end

of the rod to have a smaller wave motion than the other.

Since the fundamental frequency is in essence the smallest

frequency corresponding to the largest wavelength,

offsetting Lwould produce an unappropriatley sized

wavelength.

(c)Fundamental frequency of a steel rod:

( 5941 m /s) / [(2/1)(1.5m) = 1980.33Hz

(d) v / (2/2) L = 3960.67Hz --achieved by holding the rod

on one end.

INSTRUCTOR COMMENTS:

The ends of the rod are free so it can vibrate in any mode

for which the ends are antinodes. The greatest possible

wavelength is the one for which the ends are antinodes and

the middle is a node (form ANA). Since any place the rod is

held must be a node (your hand would quickly absorb the

energy of the vibration if there was any wave-associated

particle motion at that point) this ANA configuration is

possible only if you hold the rod at the middle.

Since a node-antinode distance corresponds to 1/4 wavelength

the ANA configuration consists of 1/2 wavelength. So 1/2

wavelength is 1.5 m and the wavelength is 3 m.

At propagation velocity 5941 m/s the 3 m wavelength implies

a frequency of 5941 m/s / (3 m) = 1980 cycles/sec or 1980

Hz, approx..

The first overtone occurs with antinodes at the ends and

node-antinode-node between, so the configuration is ANANA.

This corresponds to 4 quarter-wavelengths, or a full

wavelength. The resulting wavelength is 1.5 m and the

frequency is about 3760 Hz.

}THE FOLLOWING INFORMATION MIGHT OR MIGHT NOT BE RELATED TO

THIS PROBLEM:

STUDENT RESPONSES WITH INSTRUCTOR COMMENTS INSERTED

To find the amplitudes at 40, 20 and 10 cm from the left end:

The amplitudes are:

at 40 cm 0

at 20 cm .004m

at 10 cm .002828 m

I obtained my results by using the information in the

problem to write the equation of the standing wave. Since

the cosine function is maximum at 0, I substituted t=0 into

the equation and the value of x that I wanted to find the

amplitude for.

** wavelength = 192 m/2 / (240 Hz) = .8 m.

Amplitude at x is .4 cm sin(2 `pi x / .8 m) **the maximum

transverse velocity and acceleration at each of these points

are found from the equation of motion:

`omega = 2 `pi rad * 240 cycles / sec = 480 `pi rad/s.

.004 m * 480 `pi rad/s = 6 m/s approx and .004 m * ( 480 `pi

rad/s)^2 = 9000 m/s^2, approx.

.0028 m * 480 `pi rad/s = 4 m/s approx and .0028 m * ( 480

`pi rad/s)^2 = 6432 m/s^2, approx. **

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RESPONSE -->

okay

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&#

Very good work. Let me know if you have questions. &#