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Chapter 10: Problem 1363

For a particle executing \(\mathrm{S} . \mathrm{H} \mathrm{M} .\), when the potential energy of the oscillator becomes \(1 / 8\) the maximum potential energy, the displacement of the oscillator in terms of amplitude A will be........... (A) \((\mathrm{A} / \sqrt{2})\) (B) \(\\{\mathrm{A} /(2 \sqrt{2})\\}\) (C) \((\mathrm{A} / 2)\) (D) \(\\{\mathrm{A} /(3 \sqrt{2})\\}\)

Short Answer

Expert verified
The displacement of the oscillator in terms of amplitude A will be \((A / \sqrt{2})\).

Step by step solution

01

Relationship between potential energy and displacement

In simple harmonic motion (SHM), potential energy (U) is given by the formula: \[U = \frac{1}{2} kx^2\] where k is the spring constant, and x is the displacement from the equilibrium position. Also, the maximum potential energy (U_max) occurs when the displacement is at its maximum, which is equal to the amplitude (A), so: \[U_{\text{max}} = \frac{1}{2} kA^2\]
02

Determine the potential energy given in the exercise

We are given that the potential energy becomes 1/8 of the maximum potential energy. So, we have: \[U = \frac{1}{8} U_{\text{max}}\]
03

Substitute potential energy expressions and solve for displacement

Now, we substitute the expressions for U and U_max from Steps 1 and 2: \[\frac{1}{2} kx^2 = \frac{1}{8} \cdot \frac{1}{2} kA^2\] We can now cancel out the k and some constants: \[x^2 = \frac{1}{4} A^2\] Taking the square root of both sides, we obtain the displacement x in terms of amplitude A: \[x = \frac{A}{\sqrt{2}}\] So the correct answer is (A) \((A / \sqrt{2})\).

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Most popular questions from this chapter

The distance travelled by a particle performing S.H.M. during time interval equal to its periodic time is \(\ldots \ldots\) (A) A (B) \(2 \mathrm{~A}\) (C) \(4 \mathrm{~A}\) (D) Zero.

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 frequencies \((\mathrm{f} / \mathrm{f})=\ldots \ldots \ldots\) (A) \(\sqrt{\\{} \mathrm{M} /(\mathrm{m}+\mathrm{M})\\}\) (B) \(\sqrt{\\{\mathrm{m} /(\mathrm{m}+\mathrm{M})\\}}\) (C) \(\sqrt{\\{\mathrm{MA} / \mathrm{mA}}\\}\) (D) \(\sqrt{[}\\{(\mathrm{M}+\mathrm{m}) \mathrm{A}\\} / \mathrm{mA}]\)

A particle having mass \(1 \mathrm{~kg}\) is executing S.H.M. with an amplitude of \(0.01 \mathrm{~m}\) and a frequency of \(60 \mathrm{hz}\). The maximum force acting on this particle is \(\ldots \ldots . . \mathrm{N}\) (A) \(144 \pi^{2}\) (B) \(288 \pi^{2}\) (C) \(188 \pi^{2}\) (D) None of these. (A) \(x=a \sin 2 p \sqrt{(\ell / g) t}\) (B) \(x=a \cos 2 p \sqrt{(g / \ell) t}\) (C) \(\mathrm{x}=\mathrm{a} \sin \sqrt{(\mathrm{g} / \ell) \mathrm{t}}\) (D) \(\mathrm{x}=\mathrm{a} \cos \sqrt{(\mathrm{g} / \ell) \mathrm{t}}\)

A rocket is moving at a speed of \(130 \mathrm{~m} / \mathrm{s}\) towards a stationary target. While moving, it emits a wave of frequency $800 \mathrm{~Hz}$. Calculate the frequency of the sound as detected by the target. (Speed of wave \(=330 \mathrm{~m} / \mathrm{s}\) ) (A) \(1320 \mathrm{~Hz}\) (B) \(2540 \mathrm{~Hz}\) (C) \(1270 \mathrm{~Hz}\) (D) \(660 \mathrm{~Hz}\)

If the equation for a transverse wave is $\mathrm{y}=\mathrm{A} \operatorname{Sin} 2 \pi\( \)\\{(1 / \mathrm{T})-(\mathrm{x} / \lambda)\\}$. then for what wavelength will the maximum velocity of the particle be double the wave velocity ? (A) \((\pi \mathrm{A} / 4)\) (B) \((\pi \mathrm{A} / 2)\) (C) \(\pi \mathrm{A}\) (D) \(2 \pi \mathrm{A}\)
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