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Chapter 10: Problem 1394

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

Short Answer

Expert verified
The amplitude of the resultant oscillation is approximately 0.0125 m, which does not match any of the given answer choices.

Step by step solution

01

Identify the forces acting on the mass

When the mass is attached to the spring, it feels the spring force (Fs), the force due to gravity (Fg), and the normal force (Fn) from the inclined plane. As we are focusing on the oscillation of the mass in the direction of the spring's elongation on the inclined plane, the relevant forces are the spring force and the component of the gravitational force acting in that direction.
02

Establish the spring force equation

The spring force (Fs) follows Hooke's law, which states that the force exerted by the spring is proportional to its elongation (Δx) and the spring constant (k): Fs = k(Δx) In this case, θ = 30°, the spring constant k is given, the elongation Δx is 0.025 m, and the mass m = 1.5 kg.
03

Find the component of gravitational force acting along the spring

To find this component, we use the angle θ (30°) and the gravitational force acting on the mass (Fg = mg, g being the acceleration due to gravity): Fg_comp = Fg * sin(θ) Fg_comp = (1.5 kg)(9.81 m/s²) * sin(30°) Fg_comp ≈ 7.35 N
04

Calculate the spring constant using the spring force equation

Given that the spring is in equilibrium when the mass is initially attached, the spring force equals the component of the gravitational force acting along the spring: Fs = Fg_comp k(Δx) = Fg_comp k = Fg_comp/Δx k ≈ 7.35 N / 0.025 m k ≈ 294 N/m
05

Calculate the amplitude of the oscillation

The amplitude A of the oscillation can be found using the following equation for a simple harmonic motion: A = Δx * sin(θ) In our case, Δx = 0.025 m and θ = 30°, so we have: A = (0.025 m) * sin(30°) A ≈ 0.0125 m Comparing the calculated amplitude with the given options: (A) \(\frac{\pi}{7}\) ≈ 0.4488 (B) \(\frac{2\pi}{7}\) ≈ 0.8978 (C) \(\frac{\pi}{5}\) ≈ 0.6283 (D) \(\frac{2\pi}{5}\) ≈ 1.2566 None of the given options match the calculated amplitude exactly. Since the problem statement asks for the amplitude of the "resultant oscillation," it is possible that the correct answer is not provided, or there may be a rounding error.

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Most popular questions from this chapter

A tuning fork arrangement produces 4 beats/second with one fork of frequency \(288 \mathrm{~Hz}\). A little wax is applied on the unknown fork and it then produces 2 beats/s. The frequency of the unknown fork is $\ldots \ldots \ldots . \mathrm{Hz}$. (A) 286 (B) 292 (C) 294 (D) 288

The average values of potential energy and kinetic energy over a cycle for a S.H.O. will be ............ respectively. (A) \(0,(1 / 2) \mathrm{m} \omega^{2} \mathrm{~A}^{2}\) (B) \((1 / 2) \mathrm{m} \omega^{2} \mathrm{~A}^{2}, 0\) (C) $(1 / 2) \mathrm{m} \omega^{2} \mathrm{~A}^{2},(1 / 2) \mathrm{m} \omega^{2} \mathrm{~A}^{2}$ (D) $(1 / 4) \mathrm{m} \omega^{2} \mathrm{~A}^{2},(1 / 4) \mathrm{m} \omega^{2} \mathrm{~A}^{2}$

The ratio of force constants of two springs is \(1: 5\). The equal mass suspended at the free ends of both springs are performing S.H.M. If the maximum acceleration for both springs are equal, the ratio of amplitudes for both springs is \(\ldots \ldots\) (A) \((1 / \sqrt{5})\) (B) \((1 / 5)\) (C) \((5 / 1)\) (D) \((\sqrt{5} / 1)\)

Two masses \(m_{1}\) and \(m_{2}\) are attached to the two ends of a massless spring having force constant \(\mathrm{k}\). When the system is in equilibrium, if the mass \(\mathrm{m}_{1}\) is detached, then the angular frequency of mass \(m_{2}\) will be \(\ldots \ldots \ldots .\) (A) \(\sqrt{\left(\mathrm{k} / \mathrm{m}_{1}\right)}\) (B) \(\sqrt{\left(\mathrm{k} / \mathrm{m}^{2}\right)}\) (C) \(\sqrt{\left(k / m_{2}\right)+m_{1}}\) (D) \(\sqrt{\left\\{k /\left(m_{1}+m_{2}\right)\right\\}}\)

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