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A particle of unit mass is moving in a one- dimensional potential V(x) = x2 – x4. The minimum mechanical energy (in the same units as V(x)) above which the motion of the particle cannot be bounded for any given initial condition is _________.
(Specify your answer to two digits after the decimal point.)
Correct answer is between '0.24,0.26'. Can you explain this answer?
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A particle of unit mass is moving in a one- dimensional potential V(x)...
Finding the Minimum Mechanical Energy for the Particle in a One-Dimensional Potential
Introduction:
In this problem, we are given a one-dimensional potential V(x) = x^2 - x^4 and a particle of unit mass moving in this potential. We are required to find the minimum mechanical energy above which the motion of the particle cannot be bounded for any initial condition.
Solution:
To solve this problem, we can use the fact that for a bounded motion of the particle, the total energy E of the particle must be less than the maximum value of the potential energy V(x) at any point. If the total energy E is greater than the maximum value of the potential energy, the particle will escape to infinity.
The potential V(x) has two local minima at x = ±1, and a local maximum at x = 0. The maximum value of the potential energy occurs at x = 0 and is V(0) = 0. Therefore, for a bounded motion of the particle, the total energy E must be less than or equal to zero.
To find the minimum mechanical energy above which the motion of the particle cannot be bounded, we can use the conservation of energy principle. The total energy E of the particle is given by E = K + V(x), where K is the kinetic energy of the particle.
Since the mass of the particle is unit mass, the kinetic energy of the particle is given by K = (1/2)mv^2 = (1/2)v^2, where v is the velocity of the particle. Therefore, we have E = (1/2)v^2 + V(x) = (1/2)(v^2 - 2V(x)) + V(x).
We can rewrite this equation as E - V(x) = (1/2)(v^2 - 2V(x)). Since E ≤ 0, we have E - V(x) ≤ -V(x). Therefore, we can write (1/2)(v^2 - 2V(x)) ≤ -E + V(x). The left-hand side of this inequality is always non-negative, so we can write (1/2)(v^2 - 2V(x)) ≥ 0.
Substituting V(x) = x^2 - x^4, we get (1/2)(v^2 - 2x^2 + 2x^4) ≥ 0. This inequality holds for any value of x and v, so we can choose x = ±1 and v = 0 to get the minimum value of the left-hand side as -1/2. Therefore, we have -1/2 ≥ 0, which gives us the minimum value of E as Emin = V(±1) = -1/2.
Conclusion:
Thus, the minimum mechanical energy above which the motion of the particle cannot be bounded for any given initial condition is Emin = -1/2. In the same units as V(x), this value is approximately equal to 0.25. Therefore, the correct answer is between 0.24 and 0.26.
Introduction:
In this problem, we are given a one-dimensional potential V(x) = x^2 - x^4 and a particle of unit mass moving in this potential. We are required to find the minimum mechanical energy above which the motion of the particle cannot be bounded for any initial condition.
Solution:
To solve this problem, we can use the fact that for a bounded motion of the particle, the total energy E of the particle must be less than the maximum value of the potential energy V(x) at any point. If the total energy E is greater than the maximum value of the potential energy, the particle will escape to infinity.
The potential V(x) has two local minima at x = ±1, and a local maximum at x = 0. The maximum value of the potential energy occurs at x = 0 and is V(0) = 0. Therefore, for a bounded motion of the particle, the total energy E must be less than or equal to zero.
To find the minimum mechanical energy above which the motion of the particle cannot be bounded, we can use the conservation of energy principle. The total energy E of the particle is given by E = K + V(x), where K is the kinetic energy of the particle.
Since the mass of the particle is unit mass, the kinetic energy of the particle is given by K = (1/2)mv^2 = (1/2)v^2, where v is the velocity of the particle. Therefore, we have E = (1/2)v^2 + V(x) = (1/2)(v^2 - 2V(x)) + V(x).
We can rewrite this equation as E - V(x) = (1/2)(v^2 - 2V(x)). Since E ≤ 0, we have E - V(x) ≤ -V(x). Therefore, we can write (1/2)(v^2 - 2V(x)) ≤ -E + V(x). The left-hand side of this inequality is always non-negative, so we can write (1/2)(v^2 - 2V(x)) ≥ 0.
Substituting V(x) = x^2 - x^4, we get (1/2)(v^2 - 2x^2 + 2x^4) ≥ 0. This inequality holds for any value of x and v, so we can choose x = ±1 and v = 0 to get the minimum value of the left-hand side as -1/2. Therefore, we have -1/2 ≥ 0, which gives us the minimum value of E as Emin = V(±1) = -1/2.
Conclusion:
Thus, the minimum mechanical energy above which the motion of the particle cannot be bounded for any given initial condition is Emin = -1/2. In the same units as V(x), this value is approximately equal to 0.25. Therefore, the correct answer is between 0.24 and 0.26.
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A particle of unit mass is moving in a one- dimensional potential V(x) = x2 x4. The minimum mechanical energy (in the same units as V(x)) above which the motion of the particle cannot be bounded for any given initial condition is _________.(Specify your answer to two digits after the decimal point.)Correct answer is between '0.24,0.26'. Can you explain this answer?
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A particle of unit mass is moving in a one- dimensional potential V(x) = x2 x4. The minimum mechanical energy (in the same units as V(x)) above which the motion of the particle cannot be bounded for any given initial condition is _________.(Specify your answer to two digits after the decimal point.)Correct answer is between '0.24,0.26'. Can you explain this answer? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about A particle of unit mass is moving in a one- dimensional potential V(x) = x2 x4. The minimum mechanical energy (in the same units as V(x)) above which the motion of the particle cannot be bounded for any given initial condition is _________.(Specify your answer to two digits after the decimal point.)Correct answer is between '0.24,0.26'. Can you explain this answer? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A particle of unit mass is moving in a one- dimensional potential V(x) = x2 x4. The minimum mechanical energy (in the same units as V(x)) above which the motion of the particle cannot be bounded for any given initial condition is _________.(Specify your answer to two digits after the decimal point.)Correct answer is between '0.24,0.26'. Can you explain this answer?.
A particle of unit mass is moving in a one- dimensional potential V(x) = x2 x4. The minimum mechanical energy (in the same units as V(x)) above which the motion of the particle cannot be bounded for any given initial condition is _________.(Specify your answer to two digits after the decimal point.)Correct answer is between '0.24,0.26'. Can you explain this answer? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about A particle of unit mass is moving in a one- dimensional potential V(x) = x2 x4. The minimum mechanical energy (in the same units as V(x)) above which the motion of the particle cannot be bounded for any given initial condition is _________.(Specify your answer to two digits after the decimal point.)Correct answer is between '0.24,0.26'. Can you explain this answer? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A particle of unit mass is moving in a one- dimensional potential V(x) = x2 x4. The minimum mechanical energy (in the same units as V(x)) above which the motion of the particle cannot be bounded for any given initial condition is _________.(Specify your answer to two digits after the decimal point.)Correct answer is between '0.24,0.26'. Can you explain this answer?.
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