Get Help with 8 Oscillation and Electrical Problems | Expert Solutions Available

[physicszman]

Some chapter review problems in the back of the book I was having trouble with. Any help is appreciated. (Not asking for answers but directions and formulas explanations if possible) Thank you!

1) The Empire State Building in New York City oscillates in the wind with a period of 8 seconds. Do you think this period is different on weekdays, when full of people? If different, in which case is the period greater? Explain

2) You watch a distant woman driving nails into her porch at the rate of 1 stroke per second. You hear the sounds synchronized wirh the blows you see. You hear one more blow adter you see her stop hammering. How far away is she? The speed of sound is 343 m/s.

*She is 343 meteres away since she was hammering 1 stroke per second and u hear one more after you see the last stroke.

3) A human ear cannot hear sound at a frequency of 100 kHz or a sound at 102 kHz. But, if you walk into a room in which two sources are emitting sound at 100 kHz and at 102khz, you'll hear sound.

4) A pendulum clock can be approximated as a simple pendulum of length 1.00m and keeps accurate time where g = 9.83 m/s^2. In a location where g = 9.78 m/s^2, waht must be the new length of the pendulum, such that the clock continues to keep accurate time (that is that the period remains teh same)

5) Three point charges are fixed to the corners of a square such that the electrical field at the empty corner is zero. Do all the charges a) have the same sign and b) the same magnitude (but, possibly different signs)? Explain.

6) A small drop of water is suspended motionless in air by a uniform elecric field that is directed upward with a magnitude of 8480 N/C. The mass of the drop is 3.50 x 10^-9. a) Is the excess chargeon the drop positive or negative? Why? b) How many excess electrons or porotons reside on the drop?

7) The potetnital at a point in space has a certain value, not zero. Is the electric potential energy the same for every charge that is placed at that point? Explain

*It is the same. The value of the point does not matter.

8) If the speed of sound depended on its frequency, would you enjoy the music if you sat far away from the second balcony? Why not?

 

[Hurkyl]

Well, we have no idea what you've thought or tried for most of these problems, so how can we help you? Could you post your thoughts on all of them so far?

 

[Janitor]

Your answer to #2 is correct.

Keep trying on #7.

 

[Chen]

3) What's the question?

4) Do you know the formula that gives the time period of a simple pendulum?

5) Write down the equations that will give you the electric field at the empty corner, and see what kind of values for q1, q2 and q3 will give you zero.

6) What is the electric force that the drop is experiencing, as a result of the electric field? What must be the magnitute of its charge, and its sign, for the electric force to completely cancel the weight of the drop (mg)?

7) What is the expression of potential electric energy? Does it depend on the charge of the particle you examine, or not?

 

[physicszman]

3) Sorry! **A human ear cannot hear sound at a frequency of 100 kHz or a sound at 102 kHz. But, if you walk into a room in which two sources are emitting sound at 100 kHz and at 102khz, you'll hear sound. Explain why this is possible.

*Has to do with the two frequencies combining? (No idea sorry )

4) I know that the formula is:

T = 2p ([ L / g] )^1/2

so in the next one I am solving for L:

L = [(T^2)(g)]/(2p)

for the first equation i get T = 2s

L = [(2^2)(9.78m/s^2)]/(2p)

I get 6.2 m for the rope? I don't think its right it should be shorter than 1m because the gravity is weaker?


5) 2 corner charges are equal in magnitude and one is 2x the magnitude and opposite. they can not have the same sign I am guessing because it would just be more electric field? (Still not sure please explain equations)

6) a) The field of the particle is moving upward, and the force on the particle is upward therefore i think that positive charges move in the opposite direction of the field they are in...

b) q = m*g/E
q = (3.50 * 10 ^ -9mg * 9.81 m/s^2) / ( 8480 N/C)
q / 1.602 x 10^ -19 ( to get the number of protons)

252.74 excess protons.

7) U = k(q1q2/r)

hard time making a connection witht his problem



Thanks again!

 

[Janitor]

Note that in #7, the potential energy of a charge q in a potential V is q times V. So the PE is independent of q only for the case where V = 0, a case which has been ruled out in the problem statement.

 

[Janitor]

#3

Look into something called "beat frequency," which is the difference between two given frequencies.

 

[Janitor]

#4

The L/g ratio needs to stay the same for the period to stay the same. So if g becomes a smaller number, L must become smaller in the same ratio. Calling the first length L1, the first acceleration g1, the second length L2, and the second acceleration g2, you have:

L2=L1(g2/g1).

 

[Janitor]

#5

You are correct, they cannot all three have the same sign.

 

[Janitor]

#6

The charge will be positive as you wrote, and your analysis looks correct, as long as the mass you are using in your equation is in kilograms. It isn't clear to me from your statement of the question if that is the mass unit being used. You may or may not be off by a factor of 10^5, depending on exactly what is going on with mass units. Be careful there. Remember that a Newton is, by definition, a kilogram meter per second squared.

 

[Janitor]

#1.

A lot of vibration problems can be approximated by making an assumption of simple harmonic motion. The frequency of this type of vibration is proportional to the square root of k/m, where k is a constant that quantifies the stiffness of the structure, and m is the mass. The period of oscillation is proportional to the reciprocal of frequency, so period goes as square root of m/k.

The value of k depends on how the building was built, so it doesn't change when occupants move around in or out of the building. On weekdays, m will be bigger. So whatever happens to square root of m/k when m is increased is what will happen (approximately, anyway) to the period of oscillation. Can you tell what this square root does when you hold k constant and increase m?

 

[Janitor]

#8.

A trumpet generally puts out higher frequency pitches than a tuba. So when the trumpet player and the tuba player blow their notes at the same time, like they are supposed to, how would it sound to you (good or bad?) if the sound from the trumpet got to you before the sound from the tuba got to you? Your teacher is making the assumption that you are not some sort of weirdo, like John Cage was.

 

[physicszman]

Thanks for helping Janitor.

1) Since the mass is greater on weekdays the period of oscillation with be greater.


3) Thanks for the tip about beat frequency! The formula is Fb = |F1-F2|
Interference will alter the sound. Fb = |100kHz-102kHz| = 2kHz

the human ear can hear at 2kHz.


4) L2=L1(g2/g1)
L2=1.00m(9.78m/s^2/9.83m/s^2)
L2= 0.99m

The Length of the rope needs to be 0.99m in order for the pendulum to keep accurate time with a gravity of 9.78m/s^2.

6) Yes you were right, I forgot the copy down the unit which was kg of the drop.

To get the right answer I just did 252.74/10^5

I get .0025274 protons, is that corret?

7) Like you stated PE = qV and since V != 0 the potential electric energy is different for each point placed at that point.

8) wow, i like that explanation very much! If it depended on its frequency the higher pitch (trumpet) would arrive quicker than the lower (tuba) and if you sat on the balcony it would sound out of tune.

 

[Janitor]

The only one that may not be right yet is #6. In the first statement of the problem you wrote "The mass of the drop is 3.50 x 10^-9." There were no units quoted there. But in a later post you wrote "3.50 * 10 ^ -9mg" where mg might stand for milligrams, which are one-millionth of a kilogram.

So if the mass of the drop is 3.5x10^-9 in kilograms, I think the right answer is 2.52x10^7 excess protons.

That answer of mine agrees with your " 252.74/10^5" although it is more standard to go ahead and shift the decimal two places to the left and write it 2.5274X10^7.

If you write it as ".0025274 protons" your teacher will probably count it as wrong.