The equation for displacement of a particle at time \(t\) is given by the equation \(\mathrm{y}=3 \cos 2 \mathrm{t}+4 \sin 2 \mathrm{t}\). The maximum acceleration of the particle is $\ldots \ldots . . \mathrm{cm} / \mathrm{s}^{2}$. (A) 4 (B) 12 (C) 20 (D) 28

We are given the displacement equation: \(y = 3\cos(2t) + 4\sin(2t)\). The velocity equation is the first derivative of the displacement equation. So, we will differentiate \(y\) with respect to the time \(t\). \(\frac{dy}{dt} = -6\sin(2t) + 8\cos(2t)\) Now, we have the velocity equation.

The acceleration equation would be the second derivative of the displacement equation or the first derivative of the velocity equation. So, we will differentiate the velocity equation with respect to the time \(t\). \(\frac{d^2y}{dt^2} = -12\cos(2t) - 16\sin(2t)\) Now, we have the acceleration equation.

To find the maximum acceleration of the particle, we need to find the magnitude of the acceleration vector and seek its maximum value. We can use the Pythagorean theorem to calculate the magnitude of the acceleration vector: \(a(t) = \sqrt{(-12\cos(2t))^2 + (-16\sin(2t))^2}\) \(a(t) = \sqrt{144\cos^2(2t) + 256\sin^2(2t)}\) Now we need to find the maximum value of \(a(t)\). Using the trigonometric identity, \(\sin^2{θ} + \cos^2{θ} = 1\): \(a(t) = \sqrt{144(1 - \sin^2(2t)) + 256\sin^2(2t)}\) After replacing the identity, we have a quadratic expression in terms of \(\sin^2(2t)\): \(a(t) = \sqrt{144 + 112\sin^2(2t)}\) The maximum value of \(\sin^2(2t)\) will always be 1, as the sine function varies between -1 and 1. Substitute the maximum value of \(\sin^2(2t)\) into the equation: \(a(t)_{max} = \sqrt{144 + 112(1)}\) \(a(t)_{max} = \sqrt{256}\) Thus, the maximum acceleration of the particle is \(a(t)_{max} = 16 cm/s^2\). Since none of the given options matches the calculated maximum acceleration, there might be a typo in the options provided or an error in the problem statement.