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Asked by Madhu Last Modified
Answer As an experienced tutor registered on UrbanPro, I can confidently say that UrbanPro is one of the best platforms for online coaching and tuition. Now, let's delve into your physics question.
Given the potential energy function V(x)=kx22V(x)=2kx2, where kk is the force constant of the oscillator, and k=0.5 Nm−1k=0.5Nm−1. The task is to demonstrate that a particle with a total energy of 1 J, moving under this potential, must 'turn back' when it reaches x=±2x=±2 m.
To do this, we utilize the total energy of the system, which comprises kinetic energy (KK) and potential energy (VV). According to the law of conservation of energy, the total mechanical energy remains constant:
E=K+VE=K+V
Given that the total energy E=1E=1 J, we can express the kinetic energy as:
K=E−VK=E−V
Substituting the given potential energy function, we have:
K=1−0.5x22K=1−20.5x2
To find the turning points, where the particle changes direction, we need to consider the points where the kinetic energy is zero. At these points, all the energy is in the form of potential energy.
Setting K=0K=0, we get:
1−0.5x22=01−20.5x2=0
Solving this equation, we find:
0.5x22=120.5x2=1
0.5x2=20.5x2=2
x2=4x2=4
x=±2x=±2
So, the particle will turn back when it reaches x=±2x=±2 m, confirming the assertion. This analysis aligns with the graph of V(x)V(x) versus xx, indicating that at these points, the particle's total energy is entirely potential energy, signifying a change in direction of motion.
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