Physics for FSU

Sunday, 2 December 2007 - 5:27 AM EST

Name: "hopefully this helps :)"

Hey guys! I've been stealing your answers all semester so I figured I should help out this week. Here's an early Christmas present!

spider web:

part a: f=1/(2pi)*sqrt(k/m)... so k=(2*pi*f)^2*m

part b: f=1/(2pi)*sqrt(k/m)

example part a: A small fly of mass 0.123 g is caught in a spider's web. The web vibrates predominately with a frequency of 3.64 Hz. What is the value of the effective spring constant k for the web?

f=3.64Hz, m=.123g in kg!

so... k= (2*pi*3.64)^2*(1.23*10^-4)= 6.43*10^-2 N/m

example part b: At what frequency would you expect the web to vibrate if an insect of mass 0.494 g were trapped?

m=.494g in kg!, k=answer from part a!

so... f=1/(2pi)*sqrt[(6.43*10^-2)/(4.94*10^-4)]= 1.82 Hz

car springs oscillation:

K=F/x where m=mass of person*9.81! and x=sinkage in m!

and f=1/(2pi)*sqrt(k/m) where m=mass of person and car!

so... 1/(2pi)*sqrt[(F/x)/m]

example: A 76.0 kg person steps into a car of mass 2100.0 kg, causing it to sink 2.5 cm. Assuming no damping, with what frequency will the car and passenger vibrate on the springs

f=1/(2pi)*sqrt[(76*9.81/.025)/(2100+76)]= .588 Hz

mass on air track:

part a: x=Acos(2*pi*t/T) where A=x given and t=time and T=period

example part a: A 25.0kg block at rest on a horizontal frictionless air track is connected to the wall via a spring. The equilibrium position of the mass is defined to be at x=0. Somebody pushes the mass to the position x=.350m, then lets go. The mass undergoes simple harmonic motion with a period of 4.80 s. What is the position of the mass 4.032s after the mass is released

so... x=.350cos(2*pi*4.032/4.80) = .1875m

part b: amax=(2*pi*f)^2*A where f=1/T and A=x given

example part b:Consider the same mass and spring discussed in the previous problem. What is the magnitude of the maximum acceleration the mass undergoes during its motion?

amax=(2*pi*(1/4.80))^2*.350 = .60m/s^2

mass on a spring:

part a: E=.5kA^2 so... k=E/(.5*A^2) where E=work and A=meters compressed

part b: amax=kA/m... so m=kA/amax where k=answer from part a

example part a: A mass sitting on a horizontal, frictionless surface is attached to one end of a spring; the other end is fixed to a wall. 3.2 J of work is required to compressed the spring by 0.26 m. If the mass is released from rest with the spring compressed, it experiences a maximum acceleration of 22 m/s2. Find the spring constant.

k=3.2/[.5*(.26^2)] = 94.7N/m

example part b: Find the mass.

m= 94.7(.26)/22 = 1.12kg

wave equation:

x=Asinwt where w=angular acceleration

example: A 346 g mass vibrates according to the equation x = 0.392 sin (5.24 t) where x is in meters and t is in seconds. Determine the amplitude

example part a: A=amplitude straight from equation = .392m

example part b: Determine the frequency

w=2*pi*f so f=w/(2pi)

f=5.24/(2pi) = .834Hz

example part c: Determine the period

T=1/f where f=answerin part b

so... T=1/.834= 1.20s

example part d: Determine the total energy.

E=.5mv^2 where m in kg! and v(knot)=wa and w=2*pi*f and f=part b answer and A=part A answer so... E=.5*m*(2*pi*f*A)^2

so... E=.5*.346*(2*pi*.834*.392)^2= .730J

part e example: Determine the kinetic energy when x is 13.3 cm.

KE=.5mv^2 where v=v(knot)* sqrt[1-(x^2/A^2)] and x= new x in m! so.. KE=.5*m*[2*pi*f*A*sqrt[1-(x^2/A^2)]

so... KE=.5*.346*[2*pi*.834*.392*sqrt[1-(.133^2/.392^2)]= .646J

part f example: Determine the potential energy when x is 13.3 cm.

PE= E - KE where E=part d answer and KE=part e answer

PE= .730-.646 = .084J

SHM- mass on a spring:

part a example:

2mg=kx so... x=2mg/k m=mass given, g=9.81, and k= spring constant give

A massless spring with spring constant 27.2 N/m hangs vertically. A body of mass 0.27 kg is attached to its free end and then released. Assume that the spring was unstretched before the body was released. How far below the initial position does the body descend?

x= 2mg/k = 2(.27*9.81)/27.2 = .195m

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

part b example:

f=1/(2pi)*sqrt(k/m)

so... f= 1/(2pi) * sqrt(27.3/.27) = 1.60Hz

What is the amplitude of the resulting motion?

part c example:

F=kx where F=mass*gravity and k is given, so... x=F/k

x=F/k = .27(9.81)/27.2 = .0974m

conceptual: mass, period, spring constant:

(These are my answers and a friends)

False: Doubling the mass will double the frequency.
False: Tripling the spring constant will sextuple the period.
False: Doubling the mass will not change the frequency.
False: Doubling the spring constant will halve the period.
False: Halving the amplitude will quadruple the frequency.

False: Doubling the spring constant will halve the period.
True: Quadrupling the mass will halve the frequency.
True: Doubling the amplitude will not change the frequency.
False: Doubling the amplitude will double the period.
False: Tripling the spring constant will sextuple the period.

«	December 2007					»

S	M	T	W	T	F	S
						1
2	3	4	5	6	7	8
9	10	11	12	13	14	15
16	17	18	19	20	21	22
23	24	25	26	27	28	29
30	31