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 frequency of the particle is \(\ldots \ldots \mathrm{s}^{-1}\). (A) \((1 / \pi)\) (B) \(\pi\) (C) \((1 / 2 \pi)\) (D) \((\pi / 2)\)

Short answer

The frequency of the particle is \(\boxed{(A) \, \frac{1}{\pi} \, \text{s}^{-1}}\).

Step-by-step solution

Rewrite the displacement equation using the cosine identity

Let us use the identity \(\cos A \cos B + \sin A \sin B = \cos(A - B)\). Let \(A=2t\) and \(B=\alpha\). We want to find \(\alpha\) such that \(3\cos 2t + 4\sin 2t = R\cos(2t-\alpha)\), where \(R\geq 0\) The angle \(\alpha\) is given by: \(\tan \alpha = \frac{\text{coefficient of }\sin2t}{\text{coefficient of }\cos2t}\), which in our case means \(\tan \alpha = \frac{4}{3}\). Now we can find R: \(R = \sqrt{(3\cos 2t)^2 + (4\sin 2t)^2} = \sqrt{9\cos^2 2t + 16\sin^2 2t} = 5\) Thus, the displacement equation can be rewritten as: \(y = 5\cos(2t - \alpha)\)

Compare the simplified equation to the general harmonic equation

Now, let's compare our simplified equation to the general harmonic equation \(y = A \cos(2 \pi ft + \phi)\): \(5\cos(2t - \alpha) \, = A \cos(2 \pi ft + \phi)\)

Identify the frequency

Since our equation is \(y = 5\cos(2t - \alpha)\), the coefficient of \(t\) is 2. In the general harmonic equation, this is equal to \(2\pi f\): \(2 = 2 \pi f\) Solving for \(f\), we get: \(f = \frac{1}{\pi}\) Looking at the provided options, the frequency of the particle is \(\boxed{(A) \, \frac{1}{\pi} \, \text{s}^{-1}}\).