50fg for each correct answer, need these in 3 hours max

1 )A particle attached to a spring with k = 48 N/m is undergoing simple harmonic motion, and its position is described by the equation x = (6.3 m)cos(7.9t), with t measured in seconds.
(a) What is the mass of the particle?
(b) What is the period of the motion?
(c) What is the maximum speed of the particle?
(d) What is the maximum potential energy?
(e) What is the total energy?


2) A mass of 3.7 kg is released down a frictionless slope from a height of h = 2.1 m. When it reaches the bottom of the slope, it undergoes a completely inelastic collision with a mass of 1.0 kg attached to a spring of spring constant k = 4.9 kN/m as illustrated in the figure below. The combined mass then undergoes periodic motion. Calculate the following.

(a) maximum velocity of two—mass system
(b) frequency
(c) maximum amplitude
(d) period


(e) Determine expressions for position as a function of time and velocity as a function of time for the combined mass of this oscillating system.

x(t)=
v(t)=

f=-kx=ma

x_dot=6.3*7.9(-sin(7.9t)
x_dotdot=6.3*(7.9^2)(-cos(7.9t)

f=-48 [ (6.3 m)cos(7.9t)]=m[6.3*(7.9^2)(-cos(7.9t)]
m=48/(7.9^2)=0.76910751482
period= 2pi/7.9=0.7953399123seconds

|x_dot|=|6.3*7.9(-sin(7.9t]|=49.77 m/s max speed

(1/2)(48)(6.3^2)=952.56 j max potential energy and the total total

This post was edited by saber_x3 on Nov 28 2013 09:36pm

all correct ^^ sent fg

T = 2pi [sqrt(m/k)]
so

m = k[T^2 / 4pi^2]


f = omega /2pi
f = 7.9 / 2pi
f = 1.25

since f = 1/T
T = 1 / f
T = 1 /1.25
T = 0.8

so m = 48(0.8^2 / 4pi^2)
m = 0.77

3.7kg *2.1m(9.81m/s/s)=76.2237j

2.1m*9.81=(.5)v^2=6.41887m/s


3.7(6.41887)=4.7(v)
v=5.053 m/s of system

sorry no time to do atm

natural frequency, sqrt(k/m)

f= ((4.9/4.7)^.5)/2/pi=.162505 Hz, frequency

76.2237j

(1/2)(4.9)(x^2)=76.2237
x=5.5777 is max amplitude


This post was edited by saber_x3 on Nov 29 2013 12:11am