Monday

June 27, 2016
Posted by **shasha** on Thursday, March 10, 2011 at 10:03pm.

Experimentally it is found that a 6 kg weight stretches a certain spring 6 cm. If the weight is pulled 4 cm below the equilibrium position and released:

a. Set up the differential equation and associated conditions describing the motion.

b. Find the position of the weight as a function of time.

c. Find the amplitude, period, and frequency of motion

d. Determine the position, velocity and acceleration of the weight 0.5s after it has been released.

Question 2:

A 0.25 kg mass is horizontally attached to a spring with a stiffness 4 N/m. The damping constant b for the system is 1 N-sec/m. If the mass is displaced 0.5 m to the left and given an initial velocity of 1 m/sec to the left, find the equation of motion. What is the maximum displacement that the mass will attain?

- calculus -
**MathMate**, Friday, March 11, 2011 at 3:50pmQuestion 1:

The standard differential equation of motion is:

mx"+Bx'+kx=f(t) ....(1)

where

x" is the second derivative with respect to time of displacement, x

x' is the first derivative

m=mass

B=damping

k=spring constant

f(t)=applied external force

For the given case,

m=6 kg

B=0

k=mg/h=6*9.81/0.06=981 N/m

f(t)=0

Initial conditions:

x(0)=0.04 (note: we use the convention positive downwards)

x'(0)=0

So substitute the quantities to equation 1:

6x"+981x=0 .... (1a)

Auxiliary equation:

6m²+981=0

m=±12.79i => α=a=0, β=b=12.79

So solution to equation (1a)

x=C1 cos(bt)+C2 sin(bt)

Apply initial conditions:

x(0)=0.04 = C1(1)+C2(0) => C1=0.04

x'(t)=-C1*b*sin(bt)+C2*b*cos(bt)

x'(0)=0 = C1(0)+C2*b(1) => C2=0/b=0

Therefore C1=0.04, C2=0, or

x(t)=0.04cos(12.79t)

x'(t) = d(x(t))/dt = -12.79*0.04sin(12.79t)=-0.511sin(12.79t)

I will leave it to you to answer parts (c) and (d).

Q2 has been answered previously, see:

http://www.jiskha.com/display.cgi?id=1299811520 - calculus -
**shasha**, Saturday, March 12, 2011 at 6:46amthanxxxxx

- calculus -
**shasha**, Saturday, March 12, 2011 at 7:12amthanxxxxx...what is * stand for??

- calculus -
**MathMate**, Saturday, March 12, 2011 at 7:17amIf f(x) is a function, then f'(x)=dy/dx.

If the independent variable is known or understood (often either x or t), then

y'=dy/dx or dy/dt, depending on the context.

In this case, the independent variable understood is t, so x(t) is the function, x' stands for x'(t)=dx/dt, and x" stands for x"(t)=d²x/dt². - calculus -
**farah**, Monday, May 20, 2013 at 9:02ami still dont get it the answer for question 1, c and d..can u give me the answer please..