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Chapter 10: Problem 1374

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) \(\mathrm{x}=\mathrm{a} \cos \sqrt{(\mathrm{g} / \ell) \mathrm{t}}\)

Short Answer

Expert verified
The maximum force acting on the particle is \(14400\pi^2\) N, so the correct answer is (D) None of these.

Step by step solution

01

Calculate the angular frequency (ω)

To find the angular frequency, we'll use the formula: ω = 2πf Here, f is the frequency of the motion, which is given as 60 Hz. Plugging in the values, we have: ω = 2π(60) ω = 120π
02

Calculate the maximum acceleration (amax)

The maximum acceleration of the particle in S.H.M. can be found using the formula: amax = ω²A Here, A is the amplitude of motion, given as 0.01 m. Plugging in the values, we have: amax = (120π)²(0.01) amax = 14400π²
03

Calculate the maximum force (Fmax)

Using Newton's second law of motion (F = ma), we can find the maximum force acting on the particle: Fmax = m*amax Here, m is the mass of the particle, given as 1 kg. Plugging in the values, we have: Fmax = (1)(14400π²) Fmax = 14400π² N From the given options, the closest value is (A) 144π² N. However, the correct answer is actually 14400π² N, none of the options provided are exactly accurate. Thus, the answer is (D) None of these.

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Most popular questions from this chapter

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) \(\sqrt{2 f}\)

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) \(1+(7 \pi / 12) \mathrm{s}\)

A metal wire having linear mass density \(10 \mathrm{~g} / \mathrm{m}\) is passed over two supports separated by a distance of \(1 \mathrm{~m}\). The wire is kept in tension by suspending a \(10 \mathrm{~kg}\) mass. The mid point of the wire passes through a magnetic field provided by magnets and an a. c. supply having frequency \(\mathrm{n}\) is passed through the wire. If the wire starts vibrating with its resonant frequency, what is the frequency of a. c. supply? (A) \(50 \mathrm{~Hz}\) (B) \(100 \mathrm{~Hz}\) (C) \(200 \mathrm{~Hz}\) (D) \(25 \mathrm{~Hz}\)

A wire of length \(10 \mathrm{~m}\) and mass \(3 \mathrm{~kg}\) is suspended from a rigid support. The wire has uniform cross sectional area. Now a block of mass \(1 \mathrm{~kg}\) is suspended at the free end of the wire and a wave having wavelength \(0.05 \mathrm{~m}\) is produced at the lower end of the wire. What will be the wavelength of this wave when it reached the upper end of the wire? (A) \(0.12 \mathrm{~m}\) (B) \(0.18 \mathrm{~m}\) (C) \(0.14 \mathrm{~m}\) (D) \(0.10 \mathrm{~m}\)

Two sitar strings \(\mathrm{A}\) and \(\mathrm{B}\) playing the note "Dha" are slightly out of time and produce beats of frequency \(5 \mathrm{~Hz}\). The tension of the string B is slightly increased and the beat frequency is found to decrease to \(3 \mathrm{~Hz}\). What is the original frequency of \(\mathrm{B}\) if the frequency of \(\mathrm{A}\) is \(427 \mathrm{~Hz}\) ? (A) 432 (B) 422 (C) 437 (D) 417
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