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Chapter 10: Problem 1423

The equation for displacement of a particle at time \(\mathrm{t}\) is given by the equation \(\mathrm{y}=3 \cos 2 \mathrm{t}+4 \sin 2 \mathrm{t}\). The amplitude of oscillation is \(\ldots \ldots \ldots . \mathrm{cm}\). (A) 1 (B) 3 (C) 5 (D) 7

Short Answer

Expert verified
The amplitude of oscillation is 5 cm. The correct answer is (C) 5.

Step by step solution

01

Identify the sinusoidal function

The displacement equation is given as \(y = 3 \cos 2t + 4 \sin 2t\). This equation is a sum of a cosine and a sine function, which, in general, can be expressed as \(y = A \cos(2t - \delta)\), where \(A\) represents the amplitude and \(\delta\) represents the phase difference.
02

Find the amplitude of the oscillation

To find the amplitude, we use the formula \(A = \sqrt{a^2 + b^2}\), where \(a\) and \(b\) are the coefficients of the cosine and sine functions in the given equation, respectively. In our case, \(a = 3\) and \(b = 4\). So, the amplitude will be: \[A = \sqrt{(3)^2 + (4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5\] The amplitude of oscillation is 5 cm. The correct answer is (C) 5.

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Most popular questions from this chapter

For a particle executing \(\mathrm{S} . \mathrm{H} \mathrm{M} .\), when the potential energy of the oscillator becomes \(1 / 8\) the maximum potential energy, the displacement of the oscillator in terms of amplitude A will be........... (A) \((\mathrm{A} / \sqrt{2})\) (B) \(\\{\mathrm{A} /(2 \sqrt{2})\\}\) (C) \((\mathrm{A} / 2)\) (D) \(\\{\mathrm{A} /(3 \sqrt{2})\\}\)

When a mass \(\mathrm{m}\) is suspended from the free end of a massless spring having force constant \(\mathrm{k}\), its oscillates with frequency \(\mathrm{f}\). Now if the spring is divided into two equal parts and a mass $2 \mathrm{~m}$ is suspended from the end of anyone of them, it will oscillate with a frequency equal to......... (A) \(\mathrm{f}\) (B) \(2 \mathrm{f}\) (C) \((\mathrm{f} / \sqrt{2})\) (D) \(\sqrt{2 f}\)

The ratio of force constants of two springs is \(1: 5\). The equal mass suspended at the free ends of both springs are performing S.H.M. If the maximum acceleration for both springs are equal, the ratio of amplitudes for both springs is \(\ldots \ldots\) (A) \((1 / \sqrt{5})\) (B) \((1 / 5)\) (C) \((5 / 1)\) (D) \((\sqrt{5} / 1)\)

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 \(\mathrm{a}^{2}+\mathrm{b}^{2}\)

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) $(1 / 4) \mathrm{m} \omega^{2} \mathrm{~A}^{2},(1 / 4) \mathrm{m} \omega^{2} \mathrm{~A}^{2}$
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