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Chapter 10: Problem 1426

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

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

Step by step solution

01

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: \(y = 5\cos(2t - \alpha)\)
02

Compare the simplified equation to the general harmonic equation

Now, let's compare our simplified equation to the general harmonic equation \(y = A \cos(2 \pi ft + \phi)\): \(5\cos(2t - \alpha) \, = A \cos(2 \pi ft + \phi)\)
03

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 \(\boxed{(A) \, \frac{1}{\pi} \, \text{s}^{-1}}\).

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Most popular questions from this chapter

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

A person standing in a stationary lift measures the periodic time of a simple pendulum inside the lift to be equal to \(\mathrm{T}\). Now, if the lift moves along the vertically upward direction with an acceleration of $(\mathrm{g} / 3)$, then the periodic time of the lift will now be \((\mathrm{A}) \sqrt{3} \mathrm{~T}\) (B) \(\sqrt{(3 / 2) \mathrm{T}}\) (C) \((\mathrm{T} / 3)\) (D) \((\mathrm{T} / \sqrt{3})\)

If the equation for displacement of two particles executing S.H.M. is given by \(\mathrm{y}_{1}=2 \sin (10 \mathrm{t}+\theta)\) and $\mathrm{y}_{2}=3 \cos 10 \mathrm{t}$ respectively, then the phase difference between the velocity of two particles will be \(\ldots \ldots \ldots\) (A) \(-\theta\) (B) \(\theta\) (C) \(\theta-(\pi / 2)\) (D) \(\theta+(\pi / 2)\).

Two wires made up of same material are of equal lengths but their radii are in the ratio \(1: 2\). On stretching each of these two strings by the same tension, the ratio between their fundamental frequency is \(\ldots \ldots \ldots .\) (A) \(1: 2\) (B) \(2: 1\) (C) \(1: 4\) (D) \(4: 1\)

One end of a mass less spring having force constant \(\mathrm{k}\) and length \(50 \mathrm{~cm}\) is attached at the upper end of a plane inclined at an angle \(e=30^{\circ} .\) When a body of mass \(m=1.5 \mathrm{~kg}\) is attached at the lower end of the spring, the length of the spring increases by $2.5 \mathrm{~cm}$. Now, if the mass is displaced by a small amount and released, the amplitude of the resultant oscillation is ......... (A) \((\pi / 7)\) (B) \((2 \pi / 7)\) (C) \((\pi / 5)\) (D) \((2 \pi / 5)\)
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