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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

As shown in figure, two light springs having force constants \(\mathrm{k}_{1}=1.8 \mathrm{~N} \mathrm{~m}^{-1}\) and $\mathrm{k}_{2}=3.2 \mathrm{~N} \mathrm{~m}^{-1}$ and a block having mass \(\mathrm{m}=200 \mathrm{~g}\) are placed on a frictionless horizontal surface. One end of both springs are attached to rigid supports. The distance between the free ends of the spring is \(60 \mathrm{~cm}\) and the block is moving in this gap with a speed \(\mathrm{v}=120 \mathrm{~cm} \mathrm{~s}^{-1}\).What will be the periodic time of the block, between the two springs? (A) \(1+(5 \pi / 6) \mathrm{s}\) (B) \(1+(7 \pi / 6) \mathrm{s}\) (C) \(1+(5 \pi / 12) \mathrm{s}\) (D) \(1+(7 \pi / 12) \mathrm{s}\)

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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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) \(b\) (C) \(\mathrm{c}\) (D) \(\mathrm{d}\) Statement \(-1:\) Periodic time of a simple pendulum is independent of the mass of the bob. Statement \(-2:\) The restoring force does not depend on the mass of the bob. (A) a (B) \(b\) (C) \(\mathrm{c}\) (D) \(\mathrm{d}\)

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 motion of the particle is \(\ldots \ldots\) (A) Damped motion (B) Periodic motion (C) Rotational motion (D) S.H.M.
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