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Chapter 10: Problem 1369

As shown in figure, a spring attached to the ground vertically has a horizontal massless plate with a \(2 \mathrm{~kg}\) mass in it. When the spring (massless) is pressed slightly and released, the \(2 \mathrm{~kg}\) mass, starts executing S.H.M. The force constant of the spring is \(200 \mathrm{Nm}^{-1}\). For what minimum value of amplitude, will the mass loose contact with the plate? (Take \(\left.\mathrm{g}=10 \mathrm{~ms}^{-2}\right)\) (A) \(10.0 \mathrm{~cm}\) (B) \(8.0 \mathrm{~cm}\) (C) \(4.0 \mathrm{~cm}\) (D) For any value less than \(12.0 \mathrm{~cm}\).

Short Answer

Expert verified
The minimum amplitude for which the mass will lose contact with the plate is \(10 \thinspace \text{cm}\) (Option A).

Step by step solution

01

Write down the given information

We are given: Mass of the object (m) = 2 kg Spring constant (k) = 200 Nm Acceleration due to gravity (g) = 10 m/s²
02

Analyze the situation when mass loses contact with the plate

The mass will lose contact with the plate when the force exerted by the spring is equal to the gravitational force acting on the mass. That means: Spring force (Fs) = Gravitational force (Fg)
03

Analyze the spring force and gravitational force

The spring force (Fs) is given by Hooke's Law, \(F_s = kx\), where x is the displacement from the equilibrium position. The gravitational force (Fg) is given by, \(F_g = mg\), where m is the mass and g is the acceleration due to gravity.
04

Find the condition when the mass loses contact

As mentioned earlier, the mass loses contact when the spring force equals the gravitational force. Therefore, \(kx = mg\)
05

Calculate the minimum amplitude (x)

Our target is to find the minimum amplitude (x). From the equation in step 4, we can calculate x as: \( x = \frac{mg}{k} \) Now, we can plug in the given values: \( x = \frac{(2 \thinspace \text{kg})(10 \thinspace \text{m/s}^2)}{200 \thinspace \text{Nm}^{-1}} \) \( x = \frac{20}{200} \) \( x = 0.1 \thinspace \text{m} \) Since we need the answer in centimeters, we can convert the answer to centimeters: \( x = 0.1 \thinspace \text{m} \times \frac{100 \thinspace \text{cm}}{1 \thinspace \text{m}} = 10 \thinspace \text{cm} \) Thus, the minimum amplitude for which the mass will lose contact with the plate is 10 cm (Option A).

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Most popular questions from this chapter

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

When an elastic spring is given a displacement of \(10 \mathrm{~mm}\), it gains an potential energy equal to \(\mathrm{U}\). If this spring is given an additional displacement of \(10 \mathrm{~mm}\), then its potential energy will be.............. (A) \(\mathrm{U}\) (B) \(2 \mathrm{U}\) (C) \(4 \mathrm{U}\) (D) \(\mathrm{U} / 4\).

The amplitude for a S.H.M. given by the equation $\mathrm{x}=3 \sin 3 \mathrm{pt}+4 \cos 3 \mathrm{pt}\( is \)\ldots \ldots \ldots \ldots \mathrm{m}$ (A) 5 (B) 7 (C) 4 (D) \(3 .\)

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A particle executing S.H.M. has an amplitude \(\mathrm{A}\) and periodic time \(\mathrm{T}\). The minimum time required by the particle to get displaced by \((\mathrm{A} / \sqrt{2})\) from its equilibrium position is $\ldots \ldots \ldots \mathrm{s}$. (A) \(\mathrm{T}\) (B) \(\mathrm{T} / 4\) (C) \(\mathrm{T} / 8\) (D) \(\mathrm{T} / 16\)
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