91Ó°ÊÓ

We are given that the particle's velocity, v, is dependent on its position, x, and given by the equation: \(v^2 = a - bx^2\), where a and b are constants. Step 2: Kinematics Relation

To determine if the motion is simple harmonic and to find the frequency of oscillation, we need to check if the acceleration of the particle is directly proportional to the displacement from its mean position. To find acceleration, differentiate velocity with respect to time and then express acceleration in terms of position x. To differentiate v with respect to t, distinguish both sides of the equation with respect to x: \( v = \pm\sqrt{a - bx^2} \) \( \frac{d}{dt}(v) = \frac{d}{dt}(\pm\sqrt{a - bx^2}) \) Since \(\frac{dx}{dt} = v\): \( \frac{dv}{dt} = \frac{d}{dx}(a-bx^2)\frac{dx}{dt} \) Now, let's differentiate \(a - bx^2\): \( \frac{d(a - bx^2)}{dx} = 0 - 2bx \) Substitute this back into the equation for the time derivative of the velocity: \( \frac{dv}{dt} = (-2bx)v \) Since \(\frac{dv}{dt}\) represents acceleration, \(a(t) = -2bxv\). However, this is not in the form of \(a(t) = -\omega^2 x\), which is the requirement for simple harmonic motion. Thus, option (B) is not correct. Step 3: Frequency of Oscillation

Since the motion is not simple harmonic, the particle doesn't have a proper frequency of oscillation. Thus, option (C) is also not correct. Step 4: Direction of Motion

Given that \(v^2 = a - bx^2\), if the motion were to continue indefinitely in the positive x-direction, we must ensure that \(v^2\) stays a positive value. If \(x = 0\), then \(v^2 = a\). If \(a > 0\), then there exists an initial velocity and the particle moves. As x increases, the term \(bx^2\) also increases, and the maximum value of \(bx^2\) cannot be greater than a so that \(v^2\) remains positive. Thus, motion only happens up to a certain distance, x, along the positive x-direction before it stops and turns back. So, option (A) is not correct. Step 5: Total Energy of the Particle

As \(v^2\) is always positive, the equation \(v^2 = a - bx^2\) represents the total energy in terms of kinetic energy and potential energy. The total energy of the particle is: Total Energy = Kinetic Energy + Potential Energy \(m\frac{v^2}{2} = \frac{m}{2}(a - bx^2)\) Here, \(ma\) is equivalent to the total energy of the system, so option (D) is correct. The correct option is (D) The total energy of the particle is \(ma\).