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Chapter 10: Problem 1422

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 periodic time of oscillation is \(\ldots \ldots \ldots \ldots\) (A) \(2 \mathrm{~s}\) (B) \(\pi \mathrm{s}\) (C) \((\pi / 2) \mathrm{s}\) (D) \(2 \pi \mathrm{s}\)

Short Answer

Expert verified
The periodic time of oscillation for the given displacement equation is \(\pi\) seconds.

Step by step solution

01

Identify Angular Frequency

Let's first write the given equation: \[y = 3 \cos(2t) + 4 \sin(2t)\] Notice that in both terms of the equation, the argument of the cosine and sine functions is 2t, which means they oscillate at the same frequency. This is a very crucial observation because it tells us the angular frequency involved in the given equation. The angular frequency is represented by the coefficient of t, which in this case is 2.
02

Find the Period of Oscillation

Now that we have identified the angular frequency as 2, we can use it to find the period of oscillation. The formula to convert angular frequency (ω) and the period (T) is given by: \[\omega = \frac{2 \pi}{T}\] In this problem, we have \(\omega = 2\). So, we can write the equation as: \[2 = \frac{2 \pi}{T}\]
03

Solve for T

Now, let's solve for T: \[T = \frac{2 \pi}{2}\] \[T = \pi\] The periodic time of oscillation is \(\pi\) seconds.
04

Determine the Correct Answer

Now that we have calculated the period to be \(\pi\) seconds, we can compare it to the given options. The correct answer is: (B) \(\pi s\)

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Most popular questions from this chapter

Two sitar strings \(\mathrm{A}\) and \(\mathrm{B}\) playing the note "Dha" are slightly out of time and produce beats of frequency \(5 \mathrm{~Hz}\). The tension of the string B is slightly increased and the beat frequency is found to decrease to \(3 \mathrm{~Hz}\). What is the original frequency of \(\mathrm{B}\) if the frequency of \(\mathrm{A}\) is \(427 \mathrm{~Hz}\) ? (A) 432 (B) 422 (C) 437 (D) 417

The periodic time of two oscillators are \(\mathrm{T}\) and $(5 \mathrm{~T} / 4)$ respectively. Both oscillators starts their oscillation simultaneously from the midpoint of their path of motion. When the oscillator having periodic time \(\mathrm{T}\) completes one oscillation, the phase difference between the two oscillators will be \(\ldots \ldots \ldots\) (A) \(90^{\circ}\) (B) \(112^{\circ}\) (C) \(72^{\circ}\) (D) \(45^{\circ}\)

If the maximum frequency of a sound wave at room temperature is $20,000 \mathrm{~Hz}\( then its minimum wavelength will be approximately \)\ldots \ldots\left(\mathrm{v}=340 \mathrm{~ms}^{-1}\right)$ (A) \(0.2 \AA\) (B) \(5 \AA\) (C) \(5 \mathrm{~cm}\) to \(2 \mathrm{~m}\) (D) \(20 \mathrm{~mm}\)

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) \(1050 \mathrm{~Hz}\)

A block having mass \(\mathrm{M}\) is placed on a horizontal frictionless surface. This mass is attached to one end of a spring having force constant \(\mathrm{k}\). The other end of the spring is attached to a rigid wall. This system consisting of spring and mass \(\mathrm{M}\) is executing SHM with amplitude \(\mathrm{A}\) and frequency \(\mathrm{f}\). When the block is passing through the mid-point of its path of motion, a body of mass \(\mathrm{m}\) is placed on top of it, as a result of which its amplitude and frequency changes to \(\mathrm{A}^{\prime}\) and \(\mathrm{f}\). The ratio of amplitudes $\left(\mathrm{A}^{1} / \mathrm{A}\right)=\ldots \ldots \ldots$ (A) \(\sqrt{\\{}(\mathrm{M}+\mathrm{m}) / \mathrm{m}\\}\) (B) \(\sqrt{\\{m} /(\mathrm{M}+\mathrm{m})\\}\) (C) \(\sqrt{\\{} \mathrm{M} /(\mathrm{M}+\mathrm{m})\\}\) (D) \(\sqrt{\\{}(\mathrm{M}+\mathrm{m}) / \mathrm{M}\\}\)
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