Spider Web
A small fly of mass 0.120 g is caught in a spider's web. The web vibrates predominately with a frequency of 4.21 Hz. What is the value of the effective spring constant k for the web?
8.40×10-2 N/m
At what frequency would you expect the web to vibrate if an insect of mass 0.531 g were trapped?
2.00 Hz
8.40×10-2 N/m
At what frequency would you expect the web to vibrate if an insect of mass 0.531 g were trapped?
2.00 Hz
Discussion
NEW
For the 1st part, make sure you convert the mass to kg
NEW
Re: part 1 "A LOT OF MONSTER TRUCKS"
The second equation you gave works, but should the regular
f=(1/2pi)*sqrt(k/m)
work for this one? It seems like it should, but when I plug
in my numbers, I keep getting the wrong answer.
NEW
Please help
I cannot find the spring constant for part 1 and I'm on my
last try. I'm using the equation
1/f=2pi*sprt(m/k)
and my numbers are
f=3.68Hz
m=.110g=.00011kg
I get the answer of 6.04E-4Hz^2*kg and it says that it's
wrong. What am I doing wrong?
NEW
for part 2
Since
(1/f)=2pi(sqrt(m/k))
you can rearrange to get
f=1/(2pi*sqrt(m/k))
use the k you found in part one and make sure to change g
into kg
For the 1st part, make sure you convert the mass to kg
NEW
Re: part 1 "A LOT OF MONSTER TRUCKS"
The second equation you gave works, but should the regular
f=(1/2pi)*sqrt(k/m)
work for this one? It seems like it should, but when I plug
in my numbers, I keep getting the wrong answer.
NEW
Please help
I cannot find the spring constant for part 1 and I'm on my
last try. I'm using the equation
1/f=2pi*sprt(m/k)
and my numbers are
f=3.68Hz
m=.110g=.00011kg
I get the answer of 6.04E-4Hz^2*kg and it says that it's
wrong. What am I doing wrong?
NEW
for part 2
Since
(1/f)=2pi(sqrt(m/k))
you can rearrange to get
f=1/(2pi*sqrt(m/k))
use the k you found in part one and make sure to change g
into kg
Mass On A Spring
A 422 g mass vibrates according to the equation x = 0.338 sin (5.52 t) where x is in meters and t is in seconds. Determine the amplitude.
0.338 m
Determine the frequency.
0.879 Hz Determine the period.
1.14 s
Determine the total energy.
7.35×10-1 J Determine the kinetic energy when x is 10.8 cm.
6.60×10-1 J Determine the potential energy when x is 10.8 cm.
7.50×10-2 J
0.338 m
Determine the frequency.
0.879 Hz Determine the period.
1.14 s
Determine the total energy.
7.35×10-1 J Determine the kinetic energy when x is 10.8 cm.
6.60×10-1 J Determine the potential energy when x is 10.8 cm.
7.50×10-2 J
Discussion
NEW
Re: part Anonymous 1
part a: k=(2*work)/x^2
NEW
Part2
im kinda of confused on the second part of this problem
i set up a relationship
KA=ma(max)
My K=139.1
my a= 24m/s^2
and i though i could just divide me distance by 2 to get
the amplitude, and solve for mass, however i am getting an
incorrect value.
I would apprecaite this help, thanks.
NEW
part 2
im having trouble with finding the mass also... Can anyone help?
NEW
part b
F = kx
F also equals mass X acceleration
THEREFORE
ma=kx
and kx/a=m
good luck!
NEW
part b Anonymous 6
for finding the mass does anything need to be converted??
I've tried the equation with my numbers and it's not
working, if someone could help me out that would be great!
Thanks!
Re: part Anonymous 1
part a: k=(2*work)/x^2
NEW
Part2
im kinda of confused on the second part of this problem
i set up a relationship
KA=ma(max)
My K=139.1
my a= 24m/s^2
and i though i could just divide me distance by 2 to get
the amplitude, and solve for mass, however i am getting an
incorrect value.
I would apprecaite this help, thanks.
NEW
part 2
im having trouble with finding the mass also... Can anyone help?
NEW
part b
F = kx
F also equals mass X acceleration
THEREFORE
ma=kx
and kx/a=m
good luck!
NEW
part b Anonymous 6
for finding the mass does anything need to be converted??
I've tried the equation with my numbers and it's not
working, if someone could help me out that would be great!
Thanks!
Wave Equation
A 422 g mass vibrates according to the equation x = 0.338 sin (5.52 t) where x is in meters and t is in seconds. Determine the amplitude.
0.338 m
Determine the frequency.
0.879 Hz
Determine the period.
1.14 s
Determine the total energy.
7.35×10-1 J
Determine the kinetic energy when x is 10.8 cm.
6.60×10-1 J
Determine the potential energy when x is 10.8 cm.
7.50×10-2 J
0.338 m
Determine the frequency.
0.879 Hz
Determine the period.
1.14 s
Determine the total energy.
7.35×10-1 J
Determine the kinetic energy when x is 10.8 cm.
6.60×10-1 J
Determine the potential energy when x is 10.8 cm.
7.50×10-2 J
Discussion
NEW
Anonymous 1
don't know what to do with that equation, how do we find
part a?
NEW
Re:
You don't have to do any math. Look at the equations for
SHM: Position vs Time. Notice where A is.
NEW
Re: Re:
So, I figured out all of them except for the KE one. I'm
using the formula KE=1/2*m*v^2, where v=sqrt(k/m)*A, but
it's not the right answer. Am I using the wrong equation? I
know for a fact it's not the same formula as PE though...
NEW
Re: Re:
Nevermind! I figured out that since I have the PE and total
energy, and since KE + PE = total E, I can just solve for KE...
NEW
Anonymous 1
can't seem to get the last 3 questions.
NEW
Anonymous 4
how do I go about doing the first part?
NEW
Last 3 parts Anonymous 5
Can someone please explain the last 3 parts? Thank you
NEW
Re: Last 3 parts
For finding the total energy, you can find PEmax when the
object is at rest (since KE is equal to zero).
PEmax = 1/2*k*(A^2), where k is the spring constant and the
A is the amplitude found in part 1.
To find k, you can use the period: T = 2pi*sqrt(m/k), thus
k = m/[(T/2pi)^2].
Then, find the PE at your given x first, using PE = 1/2*k*
(x^2).
Lastly, total E - PE = KE
I hope that helps!
NEW
Part 1 Anonymous 4 (Sun Mar 7 11:29:40 am 2010 (EST))
How do I find the amplitude?
Anonymous 1
don't know what to do with that equation, how do we find
part a?
NEW
Re:
You don't have to do any math. Look at the equations for
SHM: Position vs Time. Notice where A is.
NEW
Re: Re:
So, I figured out all of them except for the KE one. I'm
using the formula KE=1/2*m*v^2, where v=sqrt(k/m)*A, but
it's not the right answer. Am I using the wrong equation? I
know for a fact it's not the same formula as PE though...
NEW
Re: Re:
Nevermind! I figured out that since I have the PE and total
energy, and since KE + PE = total E, I can just solve for KE...
NEW
Anonymous 1
can't seem to get the last 3 questions.
NEW
Anonymous 4
how do I go about doing the first part?
NEW
Last 3 parts Anonymous 5
Can someone please explain the last 3 parts? Thank you
NEW
Re: Last 3 parts
For finding the total energy, you can find PEmax when the
object is at rest (since KE is equal to zero).
PEmax = 1/2*k*(A^2), where k is the spring constant and the
A is the amplitude found in part 1.
To find k, you can use the period: T = 2pi*sqrt(m/k), thus
k = m/[(T/2pi)^2].
Then, find the PE at your given x first, using PE = 1/2*k*
(x^2).
Lastly, total E - PE = KE
I hope that helps!
NEW
Part 1 Anonymous 4 (Sun Mar 7 11:29:40 am 2010 (EST))
How do I find the amplitude?
SHM - Mass on Spring
A massless spring with spring constant 15.1 N/m hangs vertically. A body of mass 0.350 kg is attached to its free end and then released. Assume that the spring was outstretched before the body was released. How far below the initial position does the body descend?
45.5 cm
What is the frequency of oscillation of the resulting motion, assumed to be simple harmonic?
1.05 Hz
What is the amplitude of the resulting motion?
22.7 cm
45.5 cm
What is the frequency of oscillation of the resulting motion, assumed to be simple harmonic?
1.05 Hz
What is the amplitude of the resulting motion?
22.7 cm
Discussion
NEW
Anonymous 2
any advice on the third part? Please
NEW
Re:
the amplitude is going to be
0.5*ans of part 1
NEW
Anonymous 4
do you use f=-kx for part a?
NEW
first part
any help on part one I cant seem to get the correct answer
NEW
part one Anonymous 5
Im I working the first part out right? to find force I used
F= m*g then I used F = -kx Therefore x= F/-k am I
approaching this problem the wrong way?
NEW
Re: part one Anonymous 6
your right except you multipy mg by 2.
NEW
Re: Re: part one
I multiplied by 2 and it worked... but I don't understand
why. The equation [F = mg] changes [F = -kx] to [mg = -kx].
Why isn't [x = mg/-k]?
NEW
part a
can anyone tell what i am doing wrong?
F=2mg
F=2(.210kg*9.8m/s^2)=4.116N
F=-kx
4.116N=-21.4N/m*x
4.116N/-21.4N/m=x
x=.19m
i have tried both positive and negative and cant figure out
what is wrong
please help
NEW
part2
what equation do we use for part 2?
NEW
Re: Re: Re: part one Anonymous 6
I went to http://www.physicsforums.com/showthread.php?
t=195430 and "Astronuc" explained it pretty well.
NEW
In reply to the question of what equation to use for part 2,
that equation is
f=(1/pi)*sqrt(k/m)
where "f" is the frequency (in Hz), "k" is the
constant(units of kg/m^2, or N/m), and "m" is mass (in kg).
Regarding why the value for part a is A*2; this is because
the distance traveled from equilibrium position is
considered to be -A. The total distance traveled is double
this. The mass was dropped from a position in which the
spring had been stretched, however, the total distance from
the initial position is double that value of A. A is found
by dividing Force (m*g) by the k value provided in the question.
Hope this clears things up. :D
NEW
..in correction..
**In correction to the last post, the equation to determine
frequency is actually
f=(1/2pi)*sqrt(k/m)
I accidentally had that 1/pi is multiplied by the square
root of k/m . It should have been 1/2pi . My bad**
NEW
Re: ..in correction.
Correction to your correction
use the same equation as the spider web problem
(2pi * sqrt(m/k))^-1
answer units in Hz
this should give you the correct answer.
NEW
regarding the last post
Where did that formula come from? Just wondering.
NEW
The formula
2pi*sqrt(m/k)
is the formula for the period. This question is asking for
the frequency (in part b). To find the frequency, the
equation is
1/T (1 divided by the period)
where T is equal to [2pi*sqrt(m/k)]. This would also provide
the answer to part b.
NEW
If anyone is trying this at the last minute... know that
part a is asking for the answer in cm and NOT in m even
though the question does not specify that. Grrr!
Anonymous 2
any advice on the third part? Please
NEW
Re:
the amplitude is going to be
0.5*ans of part 1
NEW
Anonymous 4
do you use f=-kx for part a?
NEW
first part
any help on part one I cant seem to get the correct answer
NEW
part one Anonymous 5
Im I working the first part out right? to find force I used
F= m*g then I used F = -kx Therefore x= F/-k am I
approaching this problem the wrong way?
NEW
Re: part one Anonymous 6
your right except you multipy mg by 2.
NEW
Re: Re: part one
I multiplied by 2 and it worked... but I don't understand
why. The equation [F = mg] changes [F = -kx] to [mg = -kx].
Why isn't [x = mg/-k]?
NEW
part a
can anyone tell what i am doing wrong?
F=2mg
F=2(.210kg*9.8m/s^2)=4.116N
F=-kx
4.116N=-21.4N/m*x
4.116N/-21.4N/m=x
x=.19m
i have tried both positive and negative and cant figure out
what is wrong
please help
NEW
part2
what equation do we use for part 2?
NEW
Re: Re: Re: part one Anonymous 6
I went to http://www.physicsforums.com/showthread.php?
t=195430 and "Astronuc" explained it pretty well.
NEW
In reply to the question of what equation to use for part 2,
that equation is
f=(1/pi)*sqrt(k/m)
where "f" is the frequency (in Hz), "k" is the
constant(units of kg/m^2, or N/m), and "m" is mass (in kg).
Regarding why the value for part a is A*2; this is because
the distance traveled from equilibrium position is
considered to be -A. The total distance traveled is double
this. The mass was dropped from a position in which the
spring had been stretched, however, the total distance from
the initial position is double that value of A. A is found
by dividing Force (m*g) by the k value provided in the question.
Hope this clears things up. :D
NEW
..in correction..
**In correction to the last post, the equation to determine
frequency is actually
f=(1/2pi)*sqrt(k/m)
I accidentally had that 1/pi is multiplied by the square
root of k/m . It should have been 1/2pi . My bad**
NEW
Re: ..in correction.
Correction to your correction
use the same equation as the spider web problem
(2pi * sqrt(m/k))^-1
answer units in Hz
this should give you the correct answer.
NEW
regarding the last post
Where did that formula come from? Just wondering.
NEW
The formula
2pi*sqrt(m/k)
is the formula for the period. This question is asking for
the frequency (in part b). To find the frequency, the
equation is
1/T (1 divided by the period)
where T is equal to [2pi*sqrt(m/k)]. This would also provide
the answer to part b.
NEW
If anyone is trying this at the last minute... know that
part a is asking for the answer in cm and NOT in m even
though the question does not specify that. Grrr!
Conceptual - Mass, Period, Spring Constant
Consider a simple harmonic oscillator made of a mass sliding on a frictionless surface, and attached to a massless linear spring. Which of the following statements are true/false?
True: Quadrupling the spring constant will halve the period.
True: Doubling the amplitude will not change the frequency.
False: Tripling the spring constant will sextuple the period.
False: Halving the amplitude will quadruple the frequency.
True: Quadrupling the mass will double the period.
True: Quadrupling the spring constant will halve the period.
True: Doubling the amplitude will not change the frequency.
False: Tripling the spring constant will sextuple the period.
False: Halving the amplitude will quadruple the frequency.
True: Quadrupling the mass will double the period.
Discussion
NEW
Cant Figure it out
Hlp pleaaaassee
NEW
Help please
I am having trouble with this problem.
Halving the amplitude will quadruple the period.
Quadrupling the spring constant will halve the period.
Doubling the amplitude will not change the period.
Tripling the mass will sextuple the period.
Quadrupling the mass will halve the frequency.
NEW
Help! Anonymous 3
Any Help would be much appreciated!!!!
Cant Figure it out
Hlp pleaaaassee
NEW
Help please
I am having trouble with this problem.
Halving the amplitude will quadruple the period.
Quadrupling the spring constant will halve the period.
Doubling the amplitude will not change the period.
Tripling the mass will sextuple the period.
Quadrupling the mass will halve the frequency.
NEW
Help! Anonymous 3
Any Help would be much appreciated!!!!