A particle of mass 2 kg moving on a straight line under the action of ...
A particle of mass 2 kg moving on a straight line under the action of ...
Solution:
Given, mass of particle, m = 2 kg
Force acting on particle, F = (8 - 2x) N
Initial position of particle, x = 6 m
Let the displacement of particle from its initial position be x1.
So, x1 = x - 6
To find the time period, we need to find the relation between force and displacement. Let's differentiate the given force with respect to displacement.
F = (8 - 2x)
dF/dx = -2
This tells us that force is a constant and independent of displacement. Hence, the motion of particle is simple harmonic motion (SHM) with a constant angular frequency.
Angular frequency, w = sqrt(k/m)
where k is the force constant and is given by k = dF/dx
So, k = -2 N/m
Substituting the values of m and k, we get
w = sqrt(2) rad/s
Time period, T = 2*pi/w
T = 2*pi/sqrt(2)
T = pi*sqrt(2)
T = 4.44 s
Since the particle starts from rest, the displacement of the particle as a function of time is given by
x1 = A*sin(wt)
where A is the amplitude of the SHM.
To find the amplitude, we use the initial conditions. At t = 0, x1 = 0. So,
0 = A*sin(0)
A = 0
Hence, the amplitude is zero and the particle oscillates about its equilibrium position without crossing it.
Conclusion:
The time period of the particle is pi*sqrt(2) sec, which is approximately equal to 2*pi sec. The particle undergoes simple harmonic motion with a constant angular frequency and zero amplitude.
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