102k views
2 votes
The potential energy of a particle of mass in motion along the x-axis is given by U(x)=1​/2 kx², where x is in meters. The period of small oscillation (in seconds) is?

1 Answer

6 votes

Final answer:

The period of small oscillation for a particle described by the potential energy U(x)=1/2 kx² can be determined using the simple harmonic oscillator formula T = 2π√(m/k), but the values for mass (m) and spring constant (k) must be known to calculate an exact period value.

Step-by-step explanation:

The period of small oscillation for a particle in motion along the x-axis with a potential energy given by U(x) = 1/2 kx² can be determined using the formula for the period of a simple harmonic oscillator (SHO), which is T = 2π√(m/k). In this equation, m is the mass of the particle and k is the spring constant from the potential energy function.

The period T does not depend on the amplitude x, provided that the oscillations are small and the system follows Hooke's law, which is the case here.

However, neither m nor k is provided directly in the student's question, so additional information would be needed to calculate the exact period. Nonetheless, this formula gives us a direct relationship between mass, spring constant, and period of oscillation, helpful for understanding the dynamics of harmonic motion.

User Lyla
by
8.8k points