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Chapter 10: Problem 1421

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

Expert verified
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

01

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 \(\phi\).
02

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: \(\phi = \tan^{-1}(\frac{4}{3})\)
03

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

Now, we can rewrite the displacement equation with the calculated values of A and \(\phi\): \(y(t) = 5\cos(2t-\tan^{-1}(\frac{4}{3}))\)
04

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.

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Most popular questions from this chapter

A spring is attached to the center of a frictionless horizontal turn table and at the other end a body of mass \(2 \mathrm{~kg}\) is attached. The length of the spring is \(35 \mathrm{~cm}\). Now when the turn table is rotated with an angular speed of \(10 \mathrm{rad} \mathrm{s}^{-1}\), the length of the spring becomes \(40 \mathrm{~cm}\) then the force constant of the spring is $\ldots \ldots \mathrm{N} / \mathrm{m}$. (A) \(1.2 \times 10^{3}\) (B) \(1.6 \times 10^{3}\) (C) \(2.2 \times 10^{3}\) (D) \(2.6 \times 10^{3}\)

A body having mass \(5 \mathrm{~g}\) is executing S.H.M. with an amplitude of \(0.3 \mathrm{~m}\). If the periodic time of the system is $(\pi / 10) \mathrm{s}\(, then the maximum force acting on body is \)\ldots \ldots \ldots \ldots$ (A) \(0.6 \mathrm{~N}\) (B) \(0.3 \mathrm{~N}\) (C) \(6 \mathrm{~N}\) (D) \(3 \mathrm{~N}\)

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) \(\theta+(\pi / 2)\).

A simple pendulum is executing S.H.M. around point \(\mathrm{O}\) between the end points \(B\) and \(C\) with a periodic time of \(6 \mathrm{~s}\). If the distance between \(\mathrm{B}\) and \(\mathrm{C}\) is \(20 \mathrm{~cm}\) then in what time will the bob move from \(C\) to \(D\) ? Point \(D\) is at the mid-point of \(C\) and \(\mathrm{O}\). (A) \(1 \mathrm{~s}\) (B) \(2 \mathrm{~s}\) (C) \(3 \mathrm{~s}\) (D) \(4 \mathrm{~s}\)

The equation for displacement of a particle at time \(\mathrm{t}\) is given by the equation \(\mathrm{y}=3 \cos 2 \mathrm{t}+4 \sin 2 \mathrm{t}\). If the mass of the particle is \(5 \mathrm{gm}\), then the total energy of the particle is \(\ldots \ldots \ldots\) erg (A) 250 (B) 125 (C) 500 (D) 375
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