Question

If potential of an oscillation is u=kx3.Then its timeperiod

Answer 1

your question is incomplete. A complete question is -----> Potential energy of a particle with mass m is U=kx3 , where k is a positive constant. The particle is oscillating about the origin on x-axis. If the amplitude of oscillation is a, then its time period, T is
a)Proportional to a2
b) Independent of a
c)Proportional to √a
d)Proportional to 1/√a
Correct answer is option 'D'. Can you explain this answer


solution : potential energy, U = kx³

differentiating with respect to x,

dU/dx = 3kx²

we know, F = -dU/dx

so, F = -3kx²

Let amplitute of oscillation is a

then, maximum force , F = -3ka²

but we know in SHM, maximum force is given by, F = -m$\omega$²a

so, -m$\omega$²a = -3ka²

hence, $\omega$ = √(3ka/m)

now, but $\omega$ = 2π/T

so, T = $2\pi\sqrt{\frac{m}{3ka}}$

hence, it is clear that time period is directly proportional to 1/√a

therefore, option (d) is correct.