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A particle of charge 'q' and mass 'm' moves rectilinearly under the action of an electric field E = α - βx. Here, α and β are positive constants, and 'x' is the distance from the point where the particle was initially at rest.
Chose the correct option(s).
- a)The motion of the particle is oscillatory.
- b)The amplitude of the particle`s motion is α/β.
- c)The mean position of the particle is at x = α/β.
- d)The maximum acceleration of the particle is qα/m.
Correct answer is option 'A,B,C,D'. Can you explain this answer?
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Most Upvoted Answer
A particle of charge 'q' and mass 'm' moves rectilinearly under the a...
Solution:
Oscillatory Motion:
- The electric field experienced by the particle is given by E = α - βx.
- The force experienced by the particle is F = qE = q(α - βx).
- Using Newton's second law, F = ma, we get ma = q(α - βx).
- This is a second-order differential equation, which can be rewritten as a + (qβ/m)x = qα/m.
- This is a simple harmonic motion equation with angular frequency ω = √(qβ/m) and amplitude A = qα/(mω^2).
- Since ω and A are both positive, the motion of the particle is oscillatory. Hence, option A is correct.
Amplitude of Motion:
- The amplitude of the particle's motion is A = qα/(mω^2) = qα/(mqβ) = α/β.
- Hence, option B is correct.
Mean Position:
- The mean position of the particle is the position at which the net force on the particle is zero.
- At the mean position, α - βx = 0, which gives x = α/β.
- Hence, option C is correct.
Maximum Acceleration:
- The maximum acceleration of the particle occurs at the extreme positions, where the displacement is maximum.
- At the extreme positions, x = ±A = ±α/(mω^2β).
- The acceleration at the extreme positions is a = q(α - βx)/m = ±qα/(mω^2).
- Hence, the maximum acceleration of the particle is qα/m, which is independent of the position of the particle.
- Hence, option D is correct.
Conclusion:
- All four options are correct.
Oscillatory Motion:
- The electric field experienced by the particle is given by E = α - βx.
- The force experienced by the particle is F = qE = q(α - βx).
- Using Newton's second law, F = ma, we get ma = q(α - βx).
- This is a second-order differential equation, which can be rewritten as a + (qβ/m)x = qα/m.
- This is a simple harmonic motion equation with angular frequency ω = √(qβ/m) and amplitude A = qα/(mω^2).
- Since ω and A are both positive, the motion of the particle is oscillatory. Hence, option A is correct.
Amplitude of Motion:
- The amplitude of the particle's motion is A = qα/(mω^2) = qα/(mqβ) = α/β.
- Hence, option B is correct.
Mean Position:
- The mean position of the particle is the position at which the net force on the particle is zero.
- At the mean position, α - βx = 0, which gives x = α/β.
- Hence, option C is correct.
Maximum Acceleration:
- The maximum acceleration of the particle occurs at the extreme positions, where the displacement is maximum.
- At the extreme positions, x = ±A = ±α/(mω^2β).
- The acceleration at the extreme positions is a = q(α - βx)/m = ±qα/(mω^2).
- Hence, the maximum acceleration of the particle is qα/m, which is independent of the position of the particle.
- Hence, option D is correct.
Conclusion:
- All four options are correct.
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Community Answer
A particle of charge 'q' and mass 'm' moves rectilinearly under the a...
As we have, ma = qE so acceleration of the particle is a = q(α - β)/m
At mean position a = 0
Hence velocity of the particle is zero at x = 0 and x = 2α/β
We can conclude that particle oscillated between these two limits and have amplitude A = α/β
Acceleration of the particle is maximum at x = 0 and x = 2α/β , amax = qα/m
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A particle of charge 'q' and mass 'm' moves rectilinearly under the action of an electric field E = α - βx. Here, α and β are positive constants, and 'x' is the distance from the point where the particle was initially at rest.Chose the correct option(s).a)The motion of the particle is oscillatory.b)The amplitude of the particle`s motion is α/β.c)The mean position of the particle is at x = α/β.d)The maximum acceleration of the particle is qα/m.Correct answer is option 'A,B,C,D'. Can you explain this answer?
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A particle of charge 'q' and mass 'm' moves rectilinearly under the action of an electric field E = α - βx. Here, α and β are positive constants, and 'x' is the distance from the point where the particle was initially at rest.Chose the correct option(s).a)The motion of the particle is oscillatory.b)The amplitude of the particle`s motion is α/β.c)The mean position of the particle is at x = α/β.d)The maximum acceleration of the particle is qα/m.Correct answer is option 'A,B,C,D'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A particle of charge 'q' and mass 'm' moves rectilinearly under the action of an electric field E = α - βx. Here, α and β are positive constants, and 'x' is the distance from the point where the particle was initially at rest.Chose the correct option(s).a)The motion of the particle is oscillatory.b)The amplitude of the particle`s motion is α/β.c)The mean position of the particle is at x = α/β.d)The maximum acceleration of the particle is qα/m.Correct answer is option 'A,B,C,D'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A particle of charge 'q' and mass 'm' moves rectilinearly under the action of an electric field E = α - βx. Here, α and β are positive constants, and 'x' is the distance from the point where the particle was initially at rest.Chose the correct option(s).a)The motion of the particle is oscillatory.b)The amplitude of the particle`s motion is α/β.c)The mean position of the particle is at x = α/β.d)The maximum acceleration of the particle is qα/m.Correct answer is option 'A,B,C,D'. Can you explain this answer?.
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