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A particle of mass m is executing oscillations about the origin on x axis. its potential energy is U (x) = k|x|^3, where k is constant. If the amplitude of oscillation is a then it's time period T is?? *proportional to 1/root a *independent of a *proportional to root a *proportional to a^3/2 PLZZ SOMEONE ANSWER THIS QUESTION.
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A particle of mass m is executing oscillations about the origin on x a...
Introduction:
In this problem, we need to find the time period of oscillation of a particle with a potential energy function of U(x) = k|x|^3.
Solution:
To find the time period, we can use the formula:
T = 2π√(m/k)
where m is the mass of the particle and k is the spring constant.
Finding the spring constant:
Since the potential energy of the particle is U(x) = k|x|^3, we can find the force on the particle by taking the derivative of the potential energy function:
F(x) = -dU(x)/dx
F(x) = -3kx|x|
This force is a restoring force, which means that it always acts towards the equilibrium position (x=0). Therefore, we can write:
F(x) = -kx
where k = 3ka^2
Finding the time period:
Substituting the value of k in the formula for time period, we get:
T = 2π√(m/3ka^2)
Simplifying this expression, we get:
T = 2π√(m/3k) * 1/a
Therefore, the time period is proportional to 1/root(a).
Conclusion:
The time period of oscillation of a particle with a potential energy function of U(x) = k|x|^3 is proportional to 1/root(a). This means that as the amplitude of oscillation increases, the time period decreases.
In this problem, we need to find the time period of oscillation of a particle with a potential energy function of U(x) = k|x|^3.
Solution:
To find the time period, we can use the formula:
T = 2π√(m/k)
where m is the mass of the particle and k is the spring constant.
Finding the spring constant:
Since the potential energy of the particle is U(x) = k|x|^3, we can find the force on the particle by taking the derivative of the potential energy function:
F(x) = -dU(x)/dx
F(x) = -3kx|x|
This force is a restoring force, which means that it always acts towards the equilibrium position (x=0). Therefore, we can write:
F(x) = -kx
where k = 3ka^2
Finding the time period:
Substituting the value of k in the formula for time period, we get:
T = 2π√(m/3ka^2)
Simplifying this expression, we get:
T = 2π√(m/3k) * 1/a
Therefore, the time period is proportional to 1/root(a).
Conclusion:
The time period of oscillation of a particle with a potential energy function of U(x) = k|x|^3 is proportional to 1/root(a). This means that as the amplitude of oscillation increases, the time period decreases.
Community Answer
A particle of mass m is executing oscillations about the origin on x a...
P.E.=kx³..... F=-du/Dx=-3kx²....max force=-3ka²=-mw²a....w²=3ka/m.....nd w²also equals to 4π²/T².... comparing both T directly proportional to 1/√a..
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A particle of mass m is executing oscillations about the origin on x axis. its potential energy is U (x) = k|x|^3, where k is constant. If the amplitude of oscillation is a then it's time period T is?? *proportional to 1/root a *independent of a *proportional to root a *proportional to a^3/2 PLZZ SOMEONE ANSWER THIS QUESTION.
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A particle of mass m is executing oscillations about the origin on x axis. its potential energy is U (x) = k|x|^3, where k is constant. If the amplitude of oscillation is a then it's time period T is?? *proportional to 1/root a *independent of a *proportional to root a *proportional to a^3/2 PLZZ SOMEONE ANSWER THIS QUESTION. 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 mass m is executing oscillations about the origin on x axis. its potential energy is U (x) = k|x|^3, where k is constant. If the amplitude of oscillation is a then it's time period T is?? *proportional to 1/root a *independent of a *proportional to root a *proportional to a^3/2 PLZZ SOMEONE ANSWER THIS QUESTION. covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A particle of mass m is executing oscillations about the origin on x axis. its potential energy is U (x) = k|x|^3, where k is constant. If the amplitude of oscillation is a then it's time period T is?? *proportional to 1/root a *independent of a *proportional to root a *proportional to a^3/2 PLZZ SOMEONE ANSWER THIS QUESTION..
A particle of mass m is executing oscillations about the origin on x axis. its potential energy is U (x) = k|x|^3, where k is constant. If the amplitude of oscillation is a then it's time period T is?? *proportional to 1/root a *independent of a *proportional to root a *proportional to a^3/2 PLZZ SOMEONE ANSWER THIS QUESTION. 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 mass m is executing oscillations about the origin on x axis. its potential energy is U (x) = k|x|^3, where k is constant. If the amplitude of oscillation is a then it's time period T is?? *proportional to 1/root a *independent of a *proportional to root a *proportional to a^3/2 PLZZ SOMEONE ANSWER THIS QUESTION. covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A particle of mass m is executing oscillations about the origin on x axis. its potential energy is U (x) = k|x|^3, where k is constant. If the amplitude of oscillation is a then it's time period T is?? *proportional to 1/root a *independent of a *proportional to root a *proportional to a^3/2 PLZZ SOMEONE ANSWER THIS QUESTION..
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