Classical Mechanics - Period of the Motion

A particle of mass m moves in the potential above. The period of the motion when the particle has energy E is

A.

B.

C.

D.

E.

(GR9677 #93)

Solution:

Total Period: T = T_SHO + T_grav

For V = ½kx² → Simple Harmonic Oscillator (SHO)
Period of SHO, T_SHO = 2π√(k/m)

The graph shows only half of the usual SHO potential:
T_SHO = ½ × 2π√(k/m) = π√(k/m)

Total Period: T = π√(k/m) + T_grav

Answer: D


Notes:

To find T_grav with V = mgx:

E = T + V = 0 + mgx 
x = E/mg

Kinematic Equation:

x = v₀t +  ½gt²
v₀ = 0
x = ½gt² = E/mg
t²= 2E/mg²
t = T_grav = √(2E/mg²)

Since the particle has to travel from the origin to the right endpoint and then back to the origin, the total time contribution from this potential:

T_grav = 2√(2E/mg²)

The total period is, T = π√(k/m) + 2√(2E/mg²)