Kerala KEAM 5 June 2024 Solved Paper

The displacement of the particle in simple harmonic motion is given by:$$x = A \cos(\omega t)$$where:- $A$ is the amplitude,- $\omega$ is the angular frequency,- $t$ is the time.To find the velocity and acceleration, we use the following relationships:1. Velocity is the time derivative of displacement:$$v = \frac{dx}{dt} = -A \omega \sin(\omega t)$$At $t = \frac{T}{2}$, we know that the period $T$ is the time taken for one complete cycle of the motion, and $\omega T = 2\pi$. Thus:$$\sin\left(\omega \frac{T}{2}\right) = \sin\left(\pi\right) = 0$$So, the velocity at $t = \frac{T}{2}$ is:$$v = -A \omega \sin(\omega \frac{T}{2}) = 0$$2. Acceleration is the time derivative of velocity:$$a = \frac{dv}{dt} = -A \omega^2 \cos(\omega t)$$At $t = \frac{T}{2}$, we have:$$\cos\left(\omega \frac{T}{2}\right) = \cos\left(\pi\right) = -1$$Thus, the acceleration at $t = \frac{T}{2}$ is:$$a = -A \omega^2 \cos(\omega \frac{T}{2}) = A \omega^2$$Therefore, at $t = \frac{T}{2}$, the velocity is 0 and the acceleration is $A \omega^2$.Thus, the correct answer is:$$\boxed{-A, A \omega^2}$$ Quick Tip: At $t = \frac{T}{2}$, the particle in simple harmonic motion reaches the maximum displacement, where its velocity is zero, and the acceleration is directed towards the mean position.