Question

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 motion of the particle is \(\ldots \ldots\) (A) Damped motion (B) Periodic motion (C) Rotational motion (D) S.H.M.

Short answer

The motion of the particle is (B) Periodic motion, as determined by analyzing the equation \(y = 3\cos2t + 4\sin2t\) and rewriting it as \(y(t) = 5\cos(2t-\tan^{-1}(\frac{4}{3}))\), which represents a periodic cosine function.

Step-by-step solution

Rewrite the displacement equation in terms of sine and cosine functions

Let's rewrite the equation as: \(y(t) = A\cos(2t - \phi)\) Where A is the amplitude and \(\phi\) is the phase angle. To do this, we need to find the values of A and \(\phi\).

Calculate A and \(\phi\)

From the given equation \(y = 3\cos2t + 4\sin2t\), we have a set of simultaneous equations: \(A\cos\phi = 3\) \(A\sin\phi = 4\) To find the amplitude, A, take the square root of the sum of the squares: \(A = \sqrt{(3^2+4^2)} = 5\) To find the phase angle \(\phi\), take the inverse tangent of the ratio: \(\phi = \tan^{-1}(\frac{4}{3})\)

Rewrite the displacement equation with A and \(\phi\)

Now, we can rewrite the displacement equation with the calculated values of A and \(\phi\): \(y(t) = 5\cos(2t-\tan^{-1}(\frac{4}{3}))\)

Identify the motion type

The displacement equation represents a cosine function, which is a periodic function. Therefore, the motion of the particle is periodic. The correct answer is (B) Periodic motion.