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Chapter 10: Problem 1380

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) $\left\\{2 \mathrm{~K}_{0} / \mathrm{m} \omega\right\\}^{1 / 2} \sin \omega t$

Short Answer

Expert verified
The displacement of the particle is given by \(x = \left\\{\left(2 K_{0}\right) /\left(m \omega^{2}\right)\right\\}^{1 / 2} \sin \omega t\).

Step by step solution

01

Understand Kinetic and Potential Energy in SHM

In Simple Harmonic Motion, the total energy is the sum of kinetic energy (K) and potential energy (U), and it remains constant throughout the motion: \(E = K + U\) Since the particle is in SHM, we can write the potential energy as: \(U = \frac{1}{2} m \omega^2 x^2\) Where \(x\) is the displacement from the equilibrium position, \(m\) is the mass of the particle, and \(\omega\) is the angular frequency. #Step 2: Expressing total energy#
02

Express Total Energy in terms of K₀

The total energy (E) is constant, so we can express its initial value in terms of the maximum kinetic energy, K₀: \(E = K_0\) #Step 3: Substituting U and K in the Total Energy equation#
03

Substitute U and K into E equation

Now, substitute the expressions for U and K in the equation for total energy: \(K_0 = K + U\) \(K_0 = K_0 \cos^2(\omega t) + \frac{1}{2} m \omega^2 x^2\) #Step 4: Solving for displacement#
04

Solve for displacement x

Re-arrange the equation to solve for the displacement x: \(x^2 = \frac{2(K_0 - K_0 \cos^2(\omega t))}{m \omega^2}\) Now, take the square root of both sides to obtain the displacement expression: \(x = \sqrt{\frac{2 K_0}{m \omega^2}(1-\cos^2(\omega t))}\) Since we know that \(\sin^2(\omega t) = 1 - \cos^2(\omega t)\), we can rewrite the expression in terms of sin: \(x = \sqrt{\frac{2 K_0}{m \omega^2}} \sin(\omega t)\) This matches option (B). Therefore, the displacement of the particle is given by: \(x = \left\\{\left(2 K_{0}\right) /\left(m \omega^{2}\right)\right\\}^{1 / 2} \sin \omega t\)

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Most popular questions from this chapter

For the following questions, statement as well as the reason(s) are given. Each questions has four options. Select the correct option. (a) Statement \(-1\) is true, statement \(-2\) is true; statement \(-2\) is the correct explanation of statement \(-1\). (b) Statement \(-1\) is true, statement \(-2\) is true but statement \(-2\) is not the correct explanation of statement \(-1\). (c) Statement \(-1\) is true, statement \(-2\) is false (d) Statement \(-1\) is false, statement \(-2\) is true (A) a (B) \(\mathrm{b}\) (C) \(c\) (D) \(\mathrm{d}\) Statement \(-1:\) For a particle executing SHM, the amplitude and phase is decided by its initial position and initial velocity. Statement \(-2:\) In a SHM, the amplitude and phase is dependent on the restoring force. (A) a (B) \(b\) (C) \(\mathrm{c}\) (D) \(\mathrm{d}\)

A wire stretched between two rigid supports vibrates with a frequency of $45 \mathrm{~Hz}\(. If the mass of the wire is \)3.5 \times 10^{-2} \mathrm{~kg}$ and its linear mass density is \(4.0 \times 10^{-2} \mathrm{~kg} / \mathrm{m}\), what will be the tension in the wire? (A) \(212 \mathrm{~N}\) (B) \(236 \mathrm{~N}\) (C) \(248 \mathrm{~N}\) (D) \(254 \mathrm{~N}\)

Length of a steel wire is \(11 \mathrm{~m}\) and its mass is \(2.2 \mathrm{~kg}\). What should be the tension in the wire so that the speed of a transverse wave in it is equal to the speed of sound in dry air at \(20^{\circ} \mathrm{C}\) temperature? (A) \(2.31 \times 10^{4} \mathrm{~N}\) (B) \(2.25 \times 10^{4} \mathrm{~N}\) (C) \(2.06 \times 10^{4} \mathrm{~N}\) (D) \(2.56 \times 10^{4} \mathrm{~N}\)

The maximum velocity and maximum acceleration of a particle executing S.H.M. are \(1 \mathrm{~m} / \mathrm{s}\) and \(3.14 \mathrm{~m} / \mathrm{s}^{2}\) respectively. The frequency of oscillation for this particle is...... (A) \(0.5 \mathrm{~s}^{-1}\) (B) \(3.14 \mathrm{~s}^{-1}\) (C) \(0.25 \mathrm{~s}^{-1}\) (D) \(2 \mathrm{~s}^{-1}\)

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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