## 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?

At what frequency would you expect the web to vibrate if an insect of mass 0.531 g were trapped?

**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

## 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.

Determine the frequency.

Determine the total energy.

**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!

## 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.

Determine the frequency.

Determine the period.

Determine the total energy.

Determine the kinetic energy when x is 10.8 cm.

Determine the potential energy when x is 10.8 cm.

**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?

## 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?

What is the frequency of oscillation of the resulting motion, assumed to be simple harmonic?

What is the amplitude of the resulting motion?

**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

any advice on the third part? Please

the amplitude is going to be

0.5*ans of part 1

do you use f=-kx for part a?

any help on part one I cant seem to get the correct answer

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?

your right except you multipy mg by 2.

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]?

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

what equation do we use for part 2?

I went to http://www.physicsforums.com/showthread.php?

t=195430 and "Astronuc" explained it pretty well.

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

**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**

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.

Where did that formula come from? Just wondering.

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.

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.## 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!!!!