>does the period of a point mass moving in a potential well depend on its mass?
"For a simple harmonic oscillator, the period does not depend on the mass of the object. However, for a general potential well, the answer is more nuanced:
Simple harmonic oscillator: In this case, the period T is given by: T = 2π√(m/k)
where m is the mass and k is the spring constant. However, the frequency of oscillation ω = √(k/m), so T = 2π/ω. The spring constant k is typically proportional to m, making the period independent of mass.
General potential well: For a non-harmonic potential, the period usually does depend on mass. The period is given by:
T = 2∫dx/v(x)
where v(x) is the velocity as a function of position. This velocity depends on the total energy and the potential energy:
v(x) = √(2/m * (E - V(x)))
Here, the mass appears explicitly, and generally doesn't cancel out as it does in the simple harmonic case.
Would you like me to elaborate on any part of this explanation?"
>why do you say the spring constant k is typically proportional to m? a spring's force when stretched to a given length doesn't depend on what's attached to the end
"You're absolutely right, and I apologize for that error in my explanation. You are correct that the spring constant k is a property of the spring itself and does not depend on the mass attached to it."
"For a simple harmonic oscillator, the period does not depend on the mass of the object. However, for a general potential well, the answer is more nuanced:
Simple harmonic oscillator: In this case, the period T is given by: T = 2π√(m/k) where m is the mass and k is the spring constant. However, the frequency of oscillation ω = √(k/m), so T = 2π/ω. The spring constant k is typically proportional to m, making the period independent of mass.
General potential well: For a non-harmonic potential, the period usually does depend on mass. The period is given by: T = 2∫dx/v(x)
where v(x) is the velocity as a function of position. This velocity depends on the total energy and the potential energy:
v(x) = √(2/m * (E - V(x)))
Here, the mass appears explicitly, and generally doesn't cancel out as it does in the simple harmonic case.
Would you like me to elaborate on any part of this explanation?"
>why do you say the spring constant k is typically proportional to m? a spring's force when stretched to a given length doesn't depend on what's attached to the end
"You're absolutely right, and I apologize for that error in my explanation. You are correct that the spring constant k is a property of the spring itself and does not depend on the mass attached to it."