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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 bob of a simple pendulum having length ' \(\ell\) ' is displaced from its equilibrium position by an angle of \(\theta\) and released. If the velocity of the bob, while passing through its equilibrium position is \(\mathrm{v}\), then \(\mathrm{v}=\ldots \ldots \ldots\) (A) \(\sqrt{\\{2 g \ell(1-\cos \theta)\\}}\) (B) \(\sqrt{\\{2 g \ell(1+\sin \theta)\\}}\) (C) \(\sqrt{\\{2 g \ell(1-\sin \theta)\\}}\) (D) \(\sqrt{\\{2 g \ell(1+\cos \theta)\\}}\)

A string of length \(70 \mathrm{~cm}\) is stretched between two rigid supports. The resonant frequency for this string is found to be \(420 \mathrm{~Hz}\) and \(315 \mathrm{~Hz}\). If there are no resonant frequencies between these two values, then what would be the minimum resonant frequency of this string ? (A) \(10.5 \mathrm{~Hz}\) (B) \(1.05 \mathrm{~Hz}\) (C) \(105 \mathrm{~Hz}\) (D) \(1050 \mathrm{~Hz}\)

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

The distance travelled by a particle performing S.H.M. during time interval equal to its periodic time is \(\ldots \ldots\) (A) A (B) \(2 \mathrm{~A}\) (C) \(4 \mathrm{~A}\) (D) Zero.
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