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 ∝
∴ [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 FreeMore SHM Questions
A uniform cylinder of mass m and length l having area of cross-section a is suspended lengthwise with the help of a mass...The function sin2(ωt) represents...When two mutually perpendicular simple harmonic motions of same frequency, amplitude and phase are superimposed...The time period of an oscillating body executing SHM is 0.05 sec and its amplitude is 40 cm. The maximum velocity of par...The frequency of a simple pendulum is n oscillations per minute while that of another is (n + 1) oscillations per minute...