91Ó°ÊÓ

To find the time derivative of the displacement equation, we first need to differentiate it with respect to time, t. The displacement equation is given as: \(y = \sin^3 (\omega t)\) Differentiate it with respect to time, t: \(\frac{dy}{dt} = \frac{d}{dt}(\sin^3(\omega t))\) To find the derivative, we use the chain rule: if u = u(t) and v = v(u), then \(\frac{dv}{dt} = \frac{dv}{du}\frac{du}{dt}\). In this case, let u = \(\omega t\) and v = \(\sin^3(u)\):

First, differentiate \(v = \sin^3(u)\) with respect to u: \(\frac{dv}{du} = 3\sin^2(u)\cos(u)\) Next, differentiate \(u = \omega t\) with respect to t: \(\frac{du}{dt} = \omega\) Now, use the chain rule to find the time derivative of the displacement equation, \(\frac{dv}{dt}\): \(\frac{dv}{dt} = \frac{dv}{du}\frac{du}{dt}\) \(\frac{dy}{dt} = \omega \cdot 3\sin^2(\omega t)\cos(\omega t)\) The velocity equation for the particle is: \(v(t) = 3\omega\sin^2(\omega t)\cos(\omega t)\)

Now that we have the velocity equation, let's analyze the type of motion: - Non-periodic: The motion does not repeat itself at any regular interval, and the trigonometric function has no defined period. However, since the displacement function is a sine function raised to the power of 3, it is periodic with a period of \(T = 2\pi/\omega\). This means the motion is not non-periodic. - Periodic but not simple harmonic: For a motion to be simple harmonic, its velocity equation should be proportional to the negative of the displacement equation. In this case, the velocity equation is not proportional to the negative of the displacement equation. Hence, the motion is periodic but not simple harmonic. - Simple harmonic: As we already established, the motion is not simple harmonic because the velocity equation is not proportional to the negative of the displacement equation. Based on our analysis, the correct answer is: (B) periodic but not simple harmonic.