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Chapter 10: Problem 1337

If the equation for a particle performing S.H.M. is given by $\mathrm{y}=\sin 2 \mathrm{t}+\sqrt{3} \cos 2 \mathrm{t}\(, its periodic time will be \)\ldots \ldots .$ s. (A) 21 (B) \(\pi\) (C) \(2 \pi\) (D) \(4 \pi\).

Short Answer

Expert verified
The periodic time of the given S.H.M. is \(2\pi\) seconds. The correct answer is (C) \(2\pi\).

Step by step solution

01

Determine the period of each term

In the given equation, y(t) = sin(2t) + √3 * cos(2t). The two individual trigonometric functions are sin(2t) and cos(2t). Both of these functions have a period of \(2\pi\), because any multiple of their argument will give the same value and they only repeat at those values. For example: sin(x) = sin(x+2π) cos(x) = cos(x+2π)
02

Determine the period of their sum

To find the period of the sum, we're looking for the smallest positive value of t for which both sin(2t) and cos(2t) repeat. Since both functions have a period of \(2\pi\), their sum will also have the same period. So, the period of the given equation y(t) = sin(2t) + √3 * cos(2t) is \(2\pi\).
03

Convert the period to periodic time

The periodic time is the time taken to complete one oscillation or cycle. It is represented by the variable T. As per the given equation, when the period of the S.H.M. is \(2\pi\), the periodic time T is also \(2\pi\). So the periodic time of the given S.H.M. is \(2\pi\) seconds. The correct answer is (C) \(2\pi\).

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Most popular questions from this chapter

A wire having length \(\mathrm{L}\) is kept under tension between \(\mathrm{x}=0\) and \(\mathrm{x}=\mathrm{L}\). In one experiment, the equation of the wave and energy is given by $\mathrm{y}_{1}=\mathrm{A} \sin (\pi \mathrm{x} / \mathrm{L}) \sin \omega \mathrm{t}\( and \)\mathrm{E}_{1}$ respectively. In another experiment, it is \(\mathrm{y}_{2}=\mathrm{A} \sin\) \(\\{(2 \pi \mathrm{x}) / \mathrm{L}\\} \sin 2 \omega \mathrm{t}\) and \(\mathrm{E}_{2}\). Then........ (A) \(E_{2}=E_{1}\) (B) \(E_{2}=2 \mathrm{E}_{1}\) (C) \(\mathrm{E}_{2}=4 \mathrm{E}_{1}\) (D) \(E_{2}=16 \mathrm{E}_{1}\)

As shown in figure, a block A having mass \(M\) is attached to one end of a massless spring. The block is on a frictionless horizontal surface and the free end of the spring is attached to a wall. Another block B having mass ' \(\mathrm{m}\) ' is placed on top of block A. Now on displacing this system horizontally and released, it executes S.H.M. What should be the maximum amplitude of oscillation so that B does not slide off A? Coefficient of static friction between the surfaces of the block's is \(\mu\). (A) \(A_{\max }=\\{(\mu \mathrm{mg}) / \mathrm{k}\\}\) (B) \(A_{\max }=[\\{\mu(m+M) g\\} / k]\) (C) \(A_{\max }=[\\{\mu(M-\mathrm{m}) g\\} / \mathrm{k}]\) (D) \(A_{\max }=[\\{2 \mu(M+m)\\} / k]\)

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

If the mass of 1 mole of air is \(29 \times 10^{-3} \mathrm{~kg}\), then the speed of sound in it at STP is $(\gamma=7 / 5) .\left\\{\mathrm{T}=273 \mathrm{~K}, \mathrm{P}=1.01 \times 10^{5} \mathrm{~Pa}\right\\}$ (A) \(270 \mathrm{~m} / \mathrm{s}\) (B) \(290 \mathrm{~m} / \mathrm{s}\) (C) \(330 \mathrm{~m} / \mathrm{s}\) (D) \(350 \mathrm{~m} / \mathrm{s}\)

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