A particle oscillates in one dimension about the equilibrium position subject to a force Fx(x) that has an associated potential energy U(x). If k is the force constant, which one of the following relations is true?

examrobotsa's picture

Submitted by examrobotsa on

Q: 78 (NDA-I/2010)
A particle oscillates in one dimension about the equilibrium position subject to a force Fx(x) that has an associated potential energy U(x). If k is the force constant, which one of the following relations is true?

question_subject: 

Science

question_exam: 

NDA-I

stats: 

0,2,5,2,2,1,2

keywords: 

{'kx2': [0, 0, 1, 0], 'potential energy': [0, 0, 1, 4], 'kx': [0, 0, 1, 0], 'particle oscillates': [0, 0, 1, 0], 'jx': [0, 0, 1, 0], 'equilibrium position': [0, 0, 1, 0], 'force': [0, 0, 0, 2]}

Option 1: ¥(jx) = — kx^2

This option suggests that the restoring force Fx is proportional to the square of the displacement x. However, in simple harmonic motion, the restoring force is directly proportional to the displacement, not its square. Therefore, this option is not correct.

Option 2: Fx(x) = —kx

This option states that the force Fx is equal to the negative of the force constant k multiplied by the displacement x. This is the correct relation for a particle oscillating in one dimension. The negative sign indicates that the force is in the opposite direction of the displacement, which is necessary for oscillatory motion.

Option 3: U(x) = -kx

This option suggests that the potential energy U is equal to the negative of the force constant k multiplied by the displacement x. However, potential energy is not directly proportional to the displacement in simple harmonic motion. Therefore, this option is not correct.

Option 4: U(x) - 1 k‘x^2 2

This equation is not related to the force or potential energy in oscillatory motion. Therefore, this option is not correct.

In conclusion, the correct relation for a particle oscillating in one dimension about

Practice this on app