Question 1:
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?
Question 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
thanxxxxx
thanxxxxx...what is * stand for??
If 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.
i still don't get it the answer for question 1, c and d..can u give me the answer please..
Question 1:
a. To set up the differential equation for the motion, we can consider Hooke's Law, which relates the force exerted by a spring to its displacement. Hooke's Law states that the force exerted by a spring is directly proportional to its displacement from the equilibrium position. The equation for Hooke's Law is F = -kx, where F is the force applied by the spring, k is the spring constant, and x is the displacement from the equilibrium position.
Since we are dealing with a weight of 6 kg, we can determine the force exerted by the weight using Newton's second law of motion, F = ma, where F is the force, m is the mass, and a is the acceleration. In this case, the acceleration is proportional to the displacement, as we are dealing with simple harmonic motion. Therefore, we can write F = -kx = ma, where a is the second derivative of x with respect to time (d²x/dt²).
Given that the weight stretches the spring by 6 cm, or 0.06 m, we can set up the equation:
-kx = ma
-k(0.06) = 6a
Here, we are assuming that the equilibrium position is at x = 0. Solving for k, we get:
k = (6a) / (0.06)
Therefore, the differential equation describing the motion is:
m(d²x/dt²) = -(6a/0.06)x
b. To find the position of the weight as a function of time, we can solve the differential equation. One way to solve this is by assuming a solution of the form x = A*cos(ωt + φ), where A is the amplitude, ω is the angular frequency, t is time, and φ is the phase constant.
Substituting this assumed solution into the differential equation, we get:
m(d²/dt²)(A*cos(ωt + φ)) = -(6a/0.06)(A*cos(ωt + φ))
Differentiating twice with respect to time, we get:
-mAω²*cos(ωt + φ) = -(6a/0.06)A*cos(ωt + φ)
Simplifying the equation, we get:
ω² = (6a/0.06)/m
ω = sqrt((6a/0.06)/m)
Since ω = 2πf, where f is the frequency, we have the formula for the frequency:
f = ω/(2π)
f = sqrt((6a/0.06)/4π²m)
c. To find the amplitude, we can use the initial conditions. The weight is pulled 4 cm below the equilibrium position, which corresponds to an initial displacement of -0.04 m. Using the equation for the assumed solution, we have:
-0.04 = A*cos(φ)
Since the cosine function has a maximum amplitude of 1, we can conclude that the amplitude A is equal to 0.04 m.
The period of motion, T, can be calculated using the formula T = 1/f, where f is the frequency. Therefore, T = 1/sqrt((6a/0.06)/4π²m).
The frequency, f, is already given in part b.
d. To determine the position, velocity, and acceleration of the weight 0.5s after it has been released, we can use the equations for simple harmonic motion:
Position: x = A*cos(ωt + φ)
Velocity: v = -A*ω*sin(ωt + φ)
Acceleration: a = -A*ω²*cos(ωt + φ)
Substituting the values obtained in parts b and c, we can find the position, velocity, and acceleration at t = 0.5s.
Question 2:
To find the equation of motion for this system, we can again use Hooke's Law, F = -kx, where F is the force, k is the spring constant, and x is the displacement from the equilibrium position.
However, in this case, we have an additional damping force due to the presence of the damping constant b. The damping force is given by F_damping = -b*dx/dt, where dx/dt is the velocity.
Therefore, the equation of motion for this system is:
m(d²x/dt²) + b(dx/dt) + kx = 0
Given that the mass is 0.25 kg, the damping constant b is 1 N-sec/m, and the spring constant k is 4 N/m, we can substitute these values into the equation:
(0.25)(d²x/dt²) + (1)(dx/dt) + (4)x = 0
To find the maximum displacement that the mass will attain, we need to solve for x.