A particle P of mass 2gms moves on the X-axis and is attracted towards origin O with a force numerically equal to 8x. If it is initially at rest at X = 10, find its position at any subsequent time assuming (a) no other force acts, (b) a damping force numerically equal to 8 times the instantaneous velocity act.
Since F=ma, a = 8x/m. So, we have
a(x) = -8/m x
v(x) = -4/m x^2 + c1
s(x) = -4/3 x^3 + c1*x + c2
v(0) = 0 and s(0) = 10, so
s(x) = -4/3 x^3 + 10
For the damping, just work with
v(x) = -4/m x^2 + 8v
and follow through to get s(x)
hsbsb
Well, well, well, isn't particle P in for a fun time! Let's see what happens.
(a) If no other force acts, the particle would just keep moving towards the origin with a force numerically equal to 8x. This means that the force acting on the particle is given by F = -8x (negative because it's attractive towards the origin).
To find the position of the particle at any subsequent time, we'll have to use Newton's second law, which states that force is equal to mass times acceleration. In this case, the force is -8x and the mass of the particle is 2 grams.
So, we have -8x = 2 * a, where 'a' is the acceleration. But wait, we also know that acceleration is the second derivative of position with respect to time, so a = d^2x/dt^2.
Substituting this into our equation, we get -8x = 2 * d^2x/dt^2. Now, we can rearrange this equation to get d^2x/dt^2 = -4x.
This is a simple second-order differential equation, and the solution to this equation is x = A * cos(2t) + B * sin(2t), where A and B are constants that we can determine using the initial conditions.
Given that the particle is initially at rest at X = 10, we know that dx/dt = 0 and x = 10 when t = 0. Substituting these values into our solution equation, we get:
0 = A * cos(0) + B * sin(0),
10 = A * cos(0) + B * sin(0).
From these two equations, we can see that A = 10 and B = 0.
Therefore, the position of the particle at any subsequent time is given by x = 10 * cos(2t).
(b) Now, let's add a little twist by introducing a damping force, which is numerically equal to 8 times the instantaneous velocity (8v). Now, the force acting on the particle is given by F = -8x - 8v.
The equation of motion then becomes -8x - 8v = 2 * a, where a = d^2x/dt^2.
To solve this equation, we need to use a technique called 'undetermined coefficients'. But hey, I don't want to get too technical and bore you more than I already have!
In simpler terms, with the additional damping force, the particle's motion becomes a bit more complex. It won't just follow a simple sinusoidal path like before. Instead, it will gradually slow down and approach the origin over time.
So, buckle up, particle P! It's going to be a wild ride towards the origin with a few twists and turns thanks to that damping force! Enjoy the show!
(a) Assuming no other force acts:
According to Newton's second law, the force acting on the particle is equal to mass times acceleration (F = ma).
The force experienced by particle P towards the origin is numerically equal to 8x.
Since the particle is attracted towards the origin, the force is directed towards the left (negative X direction) for all values of X.
Let's denote the position of the particle at any time t as X(t).
We can set up the equation of motion as follows:
ma = -8X
Since the mass of the particle is given as 2g, we can convert it to kg by dividing by 1000:
0.002a = -8X
Differentiating both sides with respect to time (t), we get:
0.002(d^2X/dt^2) = -8X
Dividing by 0.002 and rearranging the equation, we have:
(d^2X/dt^2) = (-8/0.002)X
Simplifying further, we get:
d^2X/dt^2 = -4000X
This is a second-order linear homogeneous ordinary differential equation.
The general solution to this equation is of the form:
X(t) = A cos(ωt) + B sin(ωt)
Where A and B are constants determined by the initial conditions and ω is the angular frequency, given by:
ω = sqrt(4000) = 20
Now we need to find the values of A and B using the initial conditions.
Given that the particle is initially at rest at X = 10, we have:
X(0) = 10
dX/dt(0) = 0
Substituting these into the general solution, we get:
A cos(0) + B sin(0) = 10
-Aω sin(0) + Bω cos(0) = 0
Simplifying these equations, we have:
A = 10
B = 0
Therefore, the position of the particle at any subsequent time t can be given by:
X(t) = 10 cos(20t)
(b) Assuming a damping force numerically equal to 8 times the instantaneous velocity acts:
In this case, the equation of motion becomes:
ma = -8X - 8v
Where v is the instantaneous velocity of the particle.
Using the same steps as before, we can divide the equation by 0.002 and differentiate with respect to time to get:
(d^2X/dt^2) = -4000X - 4000(dX/dt)
This is the same second-order linear homogeneous differential equation as before.
Therefore, the general solution is still:
X(t) = A cos(ωt) + B sin(ωt)
Using the initial condition X(0) = 10, we get:
A = 10
To find B, we differentiate X(t) with respect to time and substitute dX/dt(0) = 0:
dX/dt = -ωA sin(ωt) + Bω cos(ωt)
dX/dt(0) = -ωA sin(0) + Bω cos(0) = -Bω = 0
B = 0
Therefore, the position of the particle at any subsequent time t in the presence of damping force can be given by:
X(t) = 10 cos(20t)
To find the position of the particle at any subsequent time, we need to solve its motion equations.
(a) When no other force acts:
The force acting on the particle is numerically equal to 8x, where x is its position on the X-axis. Assuming there are no other forces acting on the particle, we can start by writing Newton's second law of motion:
F = ma
The force acting on the particle is given as 8x. Assuming the acceleration of the particle is a, we can rewrite the equation as:
8x = 2a
Since the mass of the particle is given as 2g = 0.002kg, we substitute it in the equation:
8x = 0.002a
Now, we can rearrange the equation to solve for acceleration:
a = 4000x
The above equation relates the acceleration of the particle to its position x. To find the position at any subsequent time, we also need the initial conditions. Given that the particle is initially at rest at x = 10, we can integrate the equation to find the position as a function of time.
∫(1/a)da = ∫(1/4000x)dx
ln(a) = (1/4000)ln(x) + C1
Taking antilogarithm of both sides:
a = e^[(1/4000)ln(x) + C1]
a = e^[ln(x)^(1/4000) + C1]
a = C2*x^(1/4000)
Now we can integrate the above equation to find the velocity v as a function of time:
∫(v)dv = ∫(C2*x^(1/4000))dt
(v^2)/2 = (C2/4001)*(x^(4001/4000)) + C3
v^2 = (2C2/4001)*(x^(4001/4000)) + C4
Now we integrate again to find the position x as a function of time:
∫(dx)dx = ∫(√[(2C2/4001)*(x^(4001/4000)) + C4])dt
x = ∫(√[(2C2/4001)*(x^(4001/4000)) + C4])dt
Solving this equation requires numerical integration techniques, such as using a computer program or software like MATLAB. The position of the particle at any subsequent time can be found by evaluating this equation for the given time.
(b) When a damping force numerically equal to 8 times the instantaneous velocity acts:
In this case, the force acting on the particle will include a damping term. Newton's second law of motion can be written as:
F = ma + bv
Here, b is the damping constant and v is the velocity of the particle. Given that the damping force is numerically equal to 8 times the instantaneous velocity, we have:
F = ma - 8v
Substituting the values:
8x = 0.002a - 8v
Now, we can write the equation for acceleration using the above relation:
a = 4000x - 4000v
To solve this equation, we need to find the position x and velocity v as a function of time. The position x can be found by integrating the velocity equation:
∫(dx)dx = ∫(v)dt
x = ∫(v)dt
Since we have an equation for acceleration, we can write the equation for velocity as a function of time:
dv/dt = 4000x - 4000v
This is a first-order linear ordinary differential equation. Rearranging the terms:
dv + 4000v = 4000x dt
The general solution to this differential equation can be found using any standard method, such as the integrating factor method or separation of variables. After finding the general solution for velocity, you can integrate it again to find the position x as a function of time. However, this process is much more complex than in the case of no damping force, and the resulting equations may not have a simple analytical form. It is likely that numerical methods or software will be needed to find the particle's position at any subsequent time.