Question

A particle is oscillating as given by $U\left( y \right) = K\left| {{y^3}} \right|$ with force constant K has amplitude A. The maximum velocity during the oscillation is proportional to:
A. To A
B. Proportional to ${A^3}$
C. $\sqrt {\dfrac{{2K}}{m}{A^3}}$
D. $\sqrt {\dfrac{{2m}}{K}{A^3}}$

Answer 1

The total energy can be the maximum potential energy. Therefore, express the total energy in terms of amplitude of the oscillation. The sum of the kinetic energy and the potential energy is the total energy of the particle. The particle can have the maximum velocity if the displacement from the mean position is zero.

Formula used:
Kinetic energy, $K = \dfrac{1}{2}m{v^2}$,
where, m is the mass and v is the velocity.

Complete step by step answer:
We have given the potential of the particle, $U\left( y \right) = K\left| {{y^3}} \right|$.The particle can have the maximum velocity if its total energy is the maximum. The total energy can be the maximum potential energy. Therefore, we can express the total energy of the particle as,
$E = Ky_{\max }^2 = K{A^3}$ (Since ${y_{\max }}$ is maximum oscillation that is the amplitude of the particle)
We know that the sum of the kinetic energy and the potential energy is the total energy of the particle. Therefore, we can write,
$\dfrac{1}{2}m{v^2} + K{y^3} = K{A^3}$
$\Rightarrow \dfrac{1}{2}m{v^2} = K\left( {{A^3} - {y^3}} \right)$
$\Rightarrow {v^2} = \dfrac{{2K}}{m}\left( {{A^3} - {y^3}} \right)$
The particle can have the maximum velocity if $y = 0$ that is the mean position of the particle. Therefore, substituting $y = 0$ in the above equation, we get,
$v_{\max }^2 = \dfrac{{2K}}{m}{A^3}$
$\therefore {v_{\max }} = \sqrt {\dfrac{{2K}}{m}{A^3}}$

So, the correct answer is option C.

Note: As we know, in SHM, the potential energy of the particle is $\dfrac{1}{2}k{x^2}$, where, x is the displacement on the particle from the mean position. Since the potential energy is proportional the ${x^2}$, the total energy is proportional to ${A^2}$, where, A is the amplitude of the oscillations. In the given question, since the potential energy is proportional to ${y^3}$, the total energy should also be proportional to the ${A^3}$.