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A particle of mass 2 kg moving on a straight line under the action of force F= (8-2x) N. It is released from rest at x= 6 then time period is:- (a) pi sec (b) 4pi sec. (c) 3pi sec. (d) 2pi sec Correct option is (d). Explain with complete solution.?
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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.
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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A particle of mass 2 kg moving on a straight line under the action of force F= (8-2x) N. It is released from rest at x= 6 then time period is:- (a) pi sec (b) 4pi sec. (c) 3pi sec. (d) 2pi sec Correct option is (d). Explain with complete solution.?
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A particle of mass 2 kg moving on a straight line under the action of force F= (8-2x) N. It is released from rest at x= 6 then time period is:- (a) pi sec (b) 4pi sec. (c) 3pi sec. (d) 2pi sec Correct option is (d). Explain with complete solution.? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A particle of mass 2 kg moving on a straight line under the action of force F= (8-2x) N. It is released from rest at x= 6 then time period is:- (a) pi sec (b) 4pi sec. (c) 3pi sec. (d) 2pi sec Correct option is (d). Explain with complete solution.? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A particle of mass 2 kg moving on a straight line under the action of force F= (8-2x) N. It is released from rest at x= 6 then time period is:- (a) pi sec (b) 4pi sec. (c) 3pi sec. (d) 2pi sec Correct option is (d). Explain with complete solution.?.
A particle of mass 2 kg moving on a straight line under the action of force F= (8-2x) N. It is released from rest at x= 6 then time period is:- (a) pi sec (b) 4pi sec. (c) 3pi sec. (d) 2pi sec Correct option is (d). Explain with complete solution.? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A particle of mass 2 kg moving on a straight line under the action of force F= (8-2x) N. It is released from rest at x= 6 then time period is:- (a) pi sec (b) 4pi sec. (c) 3pi sec. (d) 2pi sec Correct option is (d). Explain with complete solution.? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A particle of mass 2 kg moving on a straight line under the action of force F= (8-2x) N. It is released from rest at x= 6 then time period is:- (a) pi sec (b) 4pi sec. (c) 3pi sec. (d) 2pi sec Correct option is (d). Explain with complete solution.?.
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