Question

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

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

Step-by-step solution

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

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)

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.

Find the condition when the mass loses contact

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

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).

Key concepts

Hooke's Law

When understanding simple harmonic motion (SHM), one key principle to grasp is Hooke's Law. This fundamental law describes how the force exerted by a spring is related to its displacement. Hooke's Law states that the force exerted by a spring is directly proportional to the amount it is stretched or compressed. It is mathematically expressed as \[F_s = kx\]where:

• \(F_s\) is the spring force,
• \(k\) is the spring constant,
• \(x\) is the displacement from the equilibrium position.

In practical terms, if you apply a force to a spring, it will stretch (or compress) by an amount proportional to that force. This relationship remains linear until the spring reaches its elastic limit. Beyond this point, Hooke's law no longer holds.By comprehending Hooke's Law, you can predict how a spring will behave under different forces, making it crucial for understanding any system involving springs.

Spring Constant

The spring constant, denoted as \(k\), is an essential factor in analyzing spring dynamics. It measures a spring's stiffness, indicating how much force is needed to stretch or compress the spring by a unit length. The larger the spring constant, the stiffer the spring and the more force required to change its length.To calculate the spring constant, you can rearrange Hooke's Law like this:\[k = \frac{F}{x}\]This equation tells us that the spring constant is the ratio of the force exerted by the spring to the displacement caused by that force.

• If \(k\) is high, even a small displacement will require a large force.
• If \(k\) is small, the spring is less stiff, and a given force will result in a larger displacement.

In real-life applications, knowing the spring constant helps engineers design systems with the desired level of elasticity. This could include anything from trampoline springs to car suspensions.

Gravitational Force

Gravitational force is a fundamental force that attracts two masses towards each other. On Earth, it is the force that pulls objects towards the center of the planet, giving them weight. The gravitational force on an object can be calculated using the formula:\[F_g = mg\]where:

• \(F_g\) is the gravitational force,
• \(m\) is the mass of the object,
• \(g\) is the acceleration due to gravity, approximately \(9.8 \thinspace \text{m/s}^2\) on Earth.

In the context of the exercise, the weight of the mass causes it to press down on the spring while at equilibrium. When a spring-mass system moves, the gravitational force interacts with the spring force. The equilibrium between these forces determines whether the mass will remain in contact or separate from the spring.Knowing how to compute gravitational force is key for solving problems involving weight and mass interactions.