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One Kinematic Problem, One Pendulum Problem, One Wave Problem

by ernay
Tags: kinematic, pendulum, wave
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ernay
#1
Feb21-09, 06:33 PM
P: 10
1. The problem statement, all variables and given/known data
ok well this was my homework over the weekend and i think i got an idea of how to solve these but im not quite sure on my answers... i've scanned the problems because they all require diagrams that must be seen to solve the problems

http://s610.photobucket.com/albums/t...=1235240535663


2. Relevant equations
problem number 1 . x = (Vi)(t)+1/2(a)(t^2)

problem number 2. . no equations necessary

problem number 3 . v = (f)(λ)




3. The attempt at a solution
problem 1 a) 1/4(g)
i couldnt figure out how to even begin with this problem so all i did was take the proportion of the string relative to each mass and then multiplied by the acceleration due to gravity
b) t = [(2h)/(1/4)(g)]
now for this one it really didnt make sense to me because could you really use this equation to find time considering the acceleration would always be changing based on its proportionality to the string... or would the answer simply be t = [(2h)/(a)]^1/2... i really didnt have an idea for this..
c) Sliding across the table to the right
d) Slides across the table to the right until it falls off the table where it'd follow a projectile path til hitting the ground
e) wasnt sure, considering the acceleration dilemma

problem 2 a) i. velocity to the right of ball, acceleration towards center
ii. acceleration down and left, no velocity

b) i. this would follow a projectile path, it'd take the motion of something falling off a cliff
ii. this would also follow a projectile path, but it'd be more parabolic?... it'd take the motion of a ball being thrown up and then later falling.

problem 3 a) λ - .6m
b) v = fλ v = (120 hz) (.6m) = 72 m/s
c) The mass should be decreased because mass is directly proportional to tension according to the formula Ft = ma. Then Tension (decreased) is directly proportional to velocity (as stated in problem). Finally velocity (decreased) is directly proportional to λ according to the formula v = fλ (the tuning forks frequency is constant). So with a shorter wavelength we will have a greater number of "loops."
d) 4cm/3 = 1.33cm
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Chi Meson
#2
Feb21-09, 07:11 PM
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Quote Quote by ernay View Post
problem 1 a) 1/4(g)
i couldnt figure out how to even begin with this problem so all i did was take the proportion of the string relative to each mass and then multiplied by the acceleration due to gravity
This makes no sense: "The proportion of the string"? Anyway, it is totally barking up the wrong tree. NOt even in the right forest. Go to Newton's 2nd Law portion of your textbook. This is a standard type of problem. THe gist is to take the total force on the system and divide it by the total mass of the system.

b) or would the answer simply be t = [(2h)/(a)]...
yes, but the "a" must be the answer to part a). ANd indicate where your equation is from. And I assumed that you meant to have a square root around those brackets.

c) Sliding across the table to the right
no point on that one if you don't describe the motion more correctly. Hint: it's not constant velocity.
d) Slides across the table to the right until it falls off the table where it'd follow a projectile path til hitting the ground
Hint: see the last hint
e) wasnt sure, considering the acceleration dilemma
What dilemma? It's a projectile. Solve for the range.
Chi Meson
#3
Feb21-09, 07:18 PM
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Quote Quote by ernay View Post
problem 3 a) λ - .6m
b) v = fλ v = (120 hz) (.6m) = 72 m/s
good start
c) The mass should be decreased because mass is directly proportional to tension according to the formula Ft = ma. Then Tension (decreased) is directly proportional to velocity (as stated in problem).
This portion is not exactly right, but it is irrelevant.
Finally velocity (decreased) is directly proportional to λ according to the formula v = fλ (the tuning forks frequency is constant). So with a shorter wavelength we will have a greater number of "loops."
This part is full points.

d) 4cm/3 = 1.33cm
No. Read the question again. Amplitude.

ernay
#4
Feb22-09, 11:39 AM
P: 10
One Kinematic Problem, One Pendulum Problem, One Wave Problem

I'm sorry but I'm still not getting 1 a...
I understood what you said but if i were to do total force divided by total mass wouldnt it simply come out to g? this is what i understood from what you said.. F/M = (2M)(g)/(2M)
where the 2M would cancel and itd simply be g? im guessing the acceleration isnt g..
I also get c and d now.. and i think e is 2h but me and my friend are unsure..

as for number 3 i appreciate the help and i see that your right... but for 3d isnt it significant that the wave starts on an antinode and then travels one whole cycle.. in essence travleing through three different antinodes, which is why i thought i would do the total vertical distance traveled divided by the number of antinodes...

and does number 2 look all right? and i dont know if it matters, but just wanted to say i use the Giancoli textbook.. oh and your right about the square root around brackets.. i just forgot to put it

I appreciate your help a lot Chi Meson! Thank You again
ernay
#5
Feb22-09, 01:55 PM
P: 10
I just confirmed my answer for number 2 on wikipedia.. but then again.. it is wikipedia.. again any help is much appreciated... 1a in particular and 3d are what i need help with
Chi Meson
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Feb22-09, 04:24 PM
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Quote Quote by ernay View Post
I'm sorry but I'm still not getting 1 a...
I understood what you said but if i were to do total force divided by total mass wouldnt it simply come out to g? this is what i understood from what you said.. F/M = (2M)(g)/(2M)
where the 2M would cancel and itd simply be g? im guessing the acceleration isnt g..
I also get c and d now.. and i think e is 2h but me and my friend are unsure..
The weight of the block on the table is balanced by the normal force there, so the total force is the weight of the hanging block, while the mass is the total mass of both blocks.


And you are "Oll Korrect" (OK) for the second question
as for number 3 i appreciate the help and i see that your right... but for 3d isnt it significant that the wave starts on an antinode and then travels one whole cycle.. in essence travleing through three different antinodes, which is why i thought i would do the total vertical distance traveled divided by the number of antinodes...
The point on the string does not move along the length of the string, just up and down as the wave moves through it. If it goes from the highest point, to the lowest point, then that is from crest to trough. What is the amplitude of a transverse wave?
ernay
#7
Feb22-09, 05:22 PM
P: 10
Ahh I now get all of it.. thank you very very much... Seems i was making a very simple mistake for 3d all along... and for 1a i now get it.. thank you very much again

Chi Meson For President! =P

EDIT : also for 1 a... the other block would have the same acceleration correct?... seems logical to me..
Chi Meson
#8
Feb22-09, 05:55 PM
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yes, this is assumed.


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