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Chapter 10: Problem 1378

The speed of a particle executing motion changes with time according to the equation $\mathrm{y}=\mathrm{a} \sin \omega \mathrm{t}+\mathrm{b} \cos \omega \mathrm{t}\(, then \)\ldots \ldots \ldots$ (A) Motion is periodic but not a S.H.M. (B) It is a S.H.M. with amplitude equal to \(\mathrm{a}+\mathrm{b}\) (C) It is a S.H.M. with amplitude equal to \(\mathrm{a}^{2}+\mathrm{b}^{2}\)

Short Answer

Expert verified
The motion of the particle is a Simple Harmonic Motion with amplitude equal to \(\sqrt{a^2 + b^2}\). Option (C) is correct.

Step by step solution

01

1. Identify the equation of the motion

First, let's observe that the given equation describes the speed of the particle, not its displacement. Simple Harmonic Motion is mainly defined by the displacement equation, so we need to find the equation for displacement. To do that, we will integrate the speed equation with respect to time.
02

2. Integrate the speed equation

The given speed equation is: \(y= a \sin(\omega t) + b \cos(\omega t)\). Now, to find the displacement equation, integrate this expression with respect to time (t): Integrating, we get: \(x(t) = -\frac{a}{\omega} \cos(\omega t) + \frac{b}{\omega} \sin(\omega t) + C\), where x(t) is the displacement and C is the integration constant.
03

3. Analyzing the displacement equation and comparing with S.H.M.

The general displacement equation for Simple Harmonic Motion is: \(x(t) = A \cos(\omega t + \phi)\) or \(x(t) = A \sin(\omega t + \phi)\), where A is the amplitude, \(\omega\) is the angular frequency, and \(\phi\) is the phase angle. Our integrated displacement equation is: \(x(t) = -\frac{a}{\omega} \cos(\omega t) + \frac{b}{\omega} \sin(\omega t) + C\). If we apply the trigonometric identity \(\cos(\alpha - \beta) = \cos(\alpha) \cos(\beta) + \sin(\alpha) \sin(\beta)\), we obtain: \(x(t) = A \cos(\omega t + \phi)\), where \(A = \sqrt{(\frac{a}{\omega})^2 + (\frac{b}{\omega})^2} = \frac{\sqrt{a^2 + b^2}}{\omega}\) and \(\phi = -\arctan(\frac{a}{b})\).
04

4. Conclusion

Since the displacement equation of the particle can be expressed in the form of a Simple Harmonic Motion equation, we can conclude that the motion is a Simple Harmonic Motion. The amplitude A can be calculated as follows: \(A = \frac{\sqrt{a^2 + b^2}}{\omega}\) Therefore, the correct option is (C) It is an S.H.M. with amplitude equal to \(\sqrt{a^2 + b^2}\).

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Most popular questions from this chapter

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) \((2 \pi / 5)\)

The displacement for a particle performing S.H.M. is given by \(\mathrm{x}=\mathrm{A} \cos (\omega \mathrm{t}+\theta)\). If the initial position of the particle is \(1 \mathrm{~cm}\) and its initial velocity is $\pi \mathrm{cms}^{-1}$, then what will be its initial phase ? The angular frequency of the particle is \(\pi \mathrm{s}^{-1}\). (A) \((2 \pi / 4)\) (B) \((7 \pi / 4)\) (C) \((5 \pi / 4)\) (D) \((3 \pi / 4)\)

When an elastic spring is given a displacement of \(10 \mathrm{~mm}\), it gains an potential energy equal to \(\mathrm{U}\). If this spring is given an additional displacement of \(10 \mathrm{~mm}\), then its potential energy will be.............. (A) \(\mathrm{U}\) (B) \(2 \mathrm{U}\) (C) \(4 \mathrm{U}\) (D) \(\mathrm{U} / 4\).

The average values of potential energy and kinetic energy over a cycle for a S.H.O. will be ............ respectively. (A) \(0,(1 / 2) \mathrm{m} \omega^{2} \mathrm{~A}^{2}\) (B) \((1 / 2) \mathrm{m} \omega^{2} \mathrm{~A}^{2}, 0\) (C) $(1 / 2) \mathrm{m} \omega^{2} \mathrm{~A}^{2},(1 / 2) \mathrm{m} \omega^{2} \mathrm{~A}^{2}$ (D) $(1 / 4) \mathrm{m} \omega^{2} \mathrm{~A}^{2},(1 / 4) \mathrm{m} \omega^{2} \mathrm{~A}^{2}$

Two simple pendulums having lengths \(144 \mathrm{~cm}\) and \(121 \mathrm{~cm}\) starts executing oscillations. At some time, both bobs of the pendulum are at the equilibrium positions and in same phase. After how many oscillations of the shorter pendulum will both the bob's pass through the equilibrium position and will have same phase ? (A) 11 (B) 12 (C) 21 (D) 20
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