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Chapter 10: Problem 1362

A particle is executing S.H.M. between \(\mathrm{x}=-\mathrm{A}\) and \(\mathrm{x}=+\mathrm{A}\). If the time taken by the particle to travel from \(\mathrm{x}=0\) to \(\mathrm{A} / 2\) is \(\mathrm{T}_{1}\) and that taken to travel from \(\mathrm{x}=\mathrm{A} / 2\) to \(\mathrm{x}=\mathrm{A}\) is \(\mathrm{T}_{2}=\) then \(\ldots .\) (A) \(\mathrm{T}_{1}<\mathrm{T}_{2}\) (B) \(\mathrm{T}_{1}>\mathrm{T}_{2}\) (C) \(\mathrm{T}_{1}=2 \mathrm{~T}_{2}\) (D) \(\mathrm{T}_{1}=\mathrm{T}_{2}\)

Short Answer

Expert verified
(A) \(T_{1}<T_{2}\)

Step by step solution

01

Review the Equation for SHM Position

In Simple Harmonic Motion, the position of a particle is given by the equation \(x(t) = A \sin{(\omega t)}\), where \(A\) is the amplitude, \(\omega\) is the angular frequency, and \(t\) is time. By knowing that the time taken to travel from an initial position to a final position can be found by comparing the corresponding sin values, we can proceed to the next step.
02

Calculate the Phase Angles for the Different Position

We need to find the phase angles that correspond to the positions \(x = 0, A/2\), and \(A\). For this, we can use the position equation of SHM. 1. For \(x = 0\), \(0 = A \sin{(\omega t_0)}\), \(\omega t_0 = 0\). 2. For \(x = A/2\), \(A/2 = A \sin{(\omega t_1)}\), \(\omega t_1 = \frac{\pi}{6}\) or \(\frac{5\pi}{6}\). 3. For \(x = A\), \(A = A \sin{(\omega t_2)}\), \(\omega t_2 = \frac{\pi}{2}\). For this exercise, we choose the smallest positive values \(\omega t_1'\) and \(\omega t_2'\) such that \(t_1\) < \(\omega t_1'\) < \(t_2\) < \( \omega t_2'\).
03

Compare T1 and T2

From the previous step, we can find the time \(T_1 = \frac{\pi}{6\omega} - \frac{0}{\omega} = \frac{\pi}{6\omega}\) and the time \(T_2 = \frac{\pi}{2\omega} - \frac{\pi}{6\omega} = \frac{\pi}{3\omega}\). Now, we can compare these times: Since \(\frac{\pi}{6\omega} < \frac{\pi}{3\omega}\), it means that \(T_1 < T_2\). Therefore, the correct answer is (A) \(\ T_{1}<\ T_{2}\).

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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) \((2 \pi / 5)\)

What should be the speed of a source of sound moving towards a stationary listener, so that the frequency of sound heard by the listener is double the frequency of sound produced by the source? \\{Speed of sound wave is \(\mathrm{v}\\}\) (A) \(\mathrm{v}\) (B) \(2 \mathrm{v}\) (C) \(\mathrm{v} / 2\) (D) \(\mathrm{v} / 4\)

As shown in figure, a spring attached to the ground vertically has a horizontal massless plate with a \(2 \mathrm{~kg}\) mass in it. When the spring (massless) is pressed slightly and released, the \(2 \mathrm{~kg}\) mass, starts executing S.H.M. The force constant of the spring is \(200 \mathrm{Nm}^{-1}\). For what minimum value of amplitude, will the mass loose contact with the plate? (Take \(\left.\mathrm{g}=10 \mathrm{~ms}^{-2}\right)\) (A) \(10.0 \mathrm{~cm}\) (B) \(8.0 \mathrm{~cm}\) (C) \(4.0 \mathrm{~cm}\) (D) For any value less than \(12.0 \mathrm{~cm}\).

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) \(\sqrt{\\{}(\mathrm{M}+\mathrm{m}) / \mathrm{M}\\}\)

The displacement for a particle performing S.H.M. is given by \(\mathrm{x}=\mathrm{A} \cos (\omega \mathrm{t}+\theta)\). If the initial position of the particle is \(1 \mathrm{~cm}\) and its initial velocity is $\pi \mathrm{cms}^{-1}$, then what will be its initial phase ? The angular frequency of the particle is \(\pi \mathrm{s}^{-1}\). (A) \((2 \pi / 4)\) (B) \((7 \pi / 4)\) (C) \((5 \pi / 4)\) (D) \((3 \pi / 4)\)
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