SHMHard

Question

A particle of mass m is executing oscillation about the origing on the x-axis. Its potential energy is U(x) = k + |x|3, where k is a positive constant. If the amplitude of oscillation is a, then its time period T is

Options

A.proportional to 1/√a
B.independent of a
C. proportional to √a
D.proportional to a3/2

Solution

U(x) = k + |x|3
∴ [k] = = = [ML-1]
Now, the time period may depend on
T ∝ (mass)x (amplitude)y (k)z
[M0L0T0] = [M]x[L]y [ML-1T-2]z = [Mx + z Ly - z T-2z]
Equating the powers, we get
- 2z = 1 or z = - 1/2
y-z = 0 or y = z = - 1/2
Hence, T ∝ (amplitude)-1/2 ∝ (a)-1/2 or T ∝

Create a free account to view solution

View Solution Free
Topic: SHM·Practice all SHM questions

More SHM Questions