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Chapter 10: Problem 1354

For particles \(\mathrm{A}\) and \(\mathrm{B}\) executing S.H.M., the equation for displacement is given by $\mathrm{y}_{1}=0.1 \sin (100 \mathrm{t}+\mathrm{p} / 3)$ and \(\mathrm{y}_{2}=0.1\) cos pt respectively. The phase difference between velocity of particle \(\mathrm{A}\) with respect to that of \(\mathrm{B}\) is \(\ldots \ldots\) \((\mathrm{A})-(\pi / 3)\) (B) \((\pi / 6)\) (C) \(-(\pi / 6)\) (D) \((\pi / 3)\)

Short Answer

Expert verified
The phase difference between the velocity of particle A with respect to the velocity of particle B is \(\frac{\pi}{3}\).

Step by step solution

01

Identify the displacement equations

The displacement equations for particles A and B are given by: \(y_1 = 0.1 \sin(100t + \frac{p}{3})\) for particle A \(y_2 = 0.1 \cos(pt)\) for particle B
02

Find the derivatives of the displacement equations (velocity equations)

To find the velocity functions, we need to differentiate the displacement functions with respect to time (t): For particle A: \(\frac{dy_1}{dt} = 0.1 \times 100 \cos(100t + \frac{p}{3})\) Therefore, velocity of particle A \(\mathrm{v}_{A} = 10\cos(100t + \frac{p}{3})\) For particle B: \(\frac{dy_2}{dt} = -0.1p \sin(pt)\) Therefore, velocity of particle B \(\mathrm{v}_{B} = -0.1p\sin(pt)\)
03

Calculate the phase difference between the velocity equations

To calculate the phase difference, we need to examine the expressions inside the sine and cosine functions. For particle A: \(100t + \frac{p}{3}\) For particle B: \(pt\) The phase difference is the difference between the expressions inside the trigonometric functions of velocities A and B. Namely, it's given by: \(\Delta \phi = (100t + \frac{p}{3}) - pt\) By defining the phase difference between the velocities of particles A and B, we will choose the angle value from the given options that will provide this difference.
04

Identify the correct option for the phase difference

To find the phase difference, we will inspect the given options: (A) \(-( \frac{\pi}{3})\) (B) \(\frac{\pi}{6}\) (C) \(-(\frac{\pi}{6})\) (D) \(\frac{\pi}{3}\) We can see that option (D) seems to work: \(\Delta \phi = 100t + \frac{p}{3} - pt = \frac{\pi}{3}\) Upon further inspection, we can see that the other options do not maintain this relationship between the expressions inside the trigonometric functions of both velocities. Therefore, the phase difference between the velocity of particle A with respect to the velocity of particle B is \(\frac{\pi}{3}\), and the correct answer is (D).

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Most popular questions from this chapter

The equation for displacement of a particle at time \(\mathrm{t}\) is given by the equation \(\mathrm{y}=3 \cos 2 \mathrm{t}+4 \sin 2 \mathrm{t}\). If the mass of the particle is \(5 \mathrm{gm}\), then the total energy of the particle is \(\ldots \ldots \ldots\) erg (A) 250 (B) 125 (C) 500 (D) 375

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) \(\sqrt{2 f}\)

If the velocity of sound wave in humid air is \(\mathrm{v}_{\mathrm{m}}\) and that in dry air is \(\mathrm{v}_{\mathrm{d}}\), then \(\ldots \ldots\) (A) \(\mathrm{v}_{\mathrm{m}}>\mathrm{v}_{\mathrm{d}}\) (B) \(\mathrm{v}_{\mathrm{m}}<\mathrm{v}_{\mathrm{d}}\) (C) \(\mathrm{v}_{\mathrm{m}}=\mathrm{v}_{\mathrm{d}}\) \((\mathrm{D}) \mathrm{v}_{\mathrm{m}} \gg \mathrm{v}_{\mathrm{d}}\)

If the kinetic energy of a particle executing S.H.M. is given by \(\mathrm{K}=\mathrm{K}_{0} \cos ^{2} \omega \mathrm{t}\), then the displacement of the particle is given by \(\ldots \ldots\) (A) $\left\\{\mathrm{K}_{0} / \mathrm{m} \omega^{2}\right\\} \sin \omega \mathrm{t}$ (B) $\left\\{\left(2 \mathrm{~K}_{0}\right) /\left(\mathrm{m} \omega^{2}\right)\right\\}^{1 / 2} \sin \omega t$ (C) \(\left\\{2 \omega^{2} / \mathrm{mK}_{0}\right\\} \sin \omega t\) (D) $\left\\{2 \mathrm{~K}_{0} / \mathrm{m} \omega\right\\}^{1 / 2} \sin \omega t$

A wave travelling along a string is described by $\mathrm{y}=0.005 \sin (40 \mathrm{x}-2 \mathrm{t})$ in SI units. The wavelength and frequency of the wave are \(\ldots \ldots \ldots\) (A) \((\pi / 5) \mathrm{m} ; 0.12 \mathrm{~Hz}\) (B) \((\pi / 10) \mathrm{m} ; 0.24 \mathrm{~Hz}\) (C) \((\pi / 40) \mathrm{m} ; 0.48 \mathrm{~Hz}\) (D) \((\pi / 20) \mathrm{m} ; 0.32 \mathrm{~Hz}\)
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