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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

A spring is attached to the center of a frictionless horizontal turn table and at the other end a body of mass \(2 \mathrm{~kg}\) is attached. The length of the spring is \(35 \mathrm{~cm}\). Now when the turn table is rotated with an angular speed of \(10 \mathrm{rad} \mathrm{s}^{-1}\), the length of the spring becomes \(40 \mathrm{~cm}\) then the force constant of the spring is $\ldots \ldots \mathrm{N} / \mathrm{m}$. (A) \(1.2 \times 10^{3}\) (B) \(1.6 \times 10^{3}\) (C) \(2.2 \times 10^{3}\) (D) \(2.6 \times 10^{3}\)

The displacement of a S.H.O. is given by the equation \(\mathrm{x}=\mathrm{A}\) \(\cos \\{\omega t+(\pi / 8)\\}\). At what time will it attain maximum velocity? (A) \((3 \pi / 8 \omega)\) (B) \((8 \pi / 3 \omega)\) (C) \((3 \pi / 16 \omega)\) (D) \((\pi / 16 \pi)\).

When a mass \(M\) is suspended from the free end of a spring, its periodic time is found to be \(\mathrm{T}\). Now, if the spring is divided into two equal parts and the same mass \(\mathrm{M}\) is suspended and oscillated, the periodic time of oscillation is found to be \(\mathrm{T}\) '. Then \(\ldots \ldots \ldots\) (A) \(\mathrm{T}<\mathrm{T}^{\prime}\) (B) \(\mathrm{T}=\mathrm{T}^{\prime}\) (C) \(\mathrm{T}>\mathrm{T}^{\prime}\) (D) Nothing can be said.

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) \(\\{\mathrm{A} /(3 \sqrt{2})\\}\)

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) \(\mathrm{U} / 4\).
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