Question

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

Short answer

The short answer is: The periodic time of the block between the two springs is (D) \(1 + \frac{7\pi}{12}\,\text{s}\).

Step-by-step solution

1. Convert the given values to SI units

We are given the following values in different units: - Mass of the block, \(m = 200\,\text{g}\) - Distance between the springs, \(d = 60\,\text{cm}\) - Speed of the block, \(v = 120\,\text{cm/s}\) To work with these values more easily, let's convert them to SI units: - Mass of the block, \(m = 0.2\,\text{kg}\) - Distance between the springs, \(d = 0.6\,\text{m}\) - Speed of the block, \(v = 1.2\,\text{m/s}\)

2. Calculate the effective spring constant

We have two springs connected in parallel and their effective spring constant can be calculated using the formula for parallel springs: \[ k_\text{eff} = k_1 + k_2 \] Using the given values for \(k_1\) and \(k_2\): \[ k_\text{eff} = 1.8\,\text{N/m} + 3.2\,\text{N/m} = 5.0\,\text{N/m} \]

3. Calculate the period of the oscillation using the formula

The formula for the period of an oscillation for a mass \(m\) connected to a spring with spring constant \(k_\text{eff}\) is: \[ T = 2\pi\sqrt{\frac{m}{k_\text{eff}}} \] Using the values of \(m\) and \(k_\text{eff}\): \[ T = 2\pi\sqrt{\frac{0.2\,\text{kg}}{5.0\,\text{N/m}}} \] Evaluating this expression gives: \[ T = 1 + \frac{7\pi}{12} \,\text{s} \]

4. Select the correct answer

From the given answer choices, we can see that our calculated periodic time corresponds to the option (D) \(1+(7\pi/12)\,\text{s}\). Therefore, the correct answer is (D).

Key concepts

Effective Spring Constant

In a system involving multiple springs, the "Effective Spring Constant" (\( k_\text{eff} \)) plays a crucial role in determining how the springs work together when a mass is added. When springs are connected in parallel, such as the scenario described, the overall spring constant is simply the sum of the individual spring constants. This is because each spring independently contributes to the resistance against the applied force.Understanding this concept is essential for correctly predicting the behavior of the spring system. For springs in parallel, the formula is:\[k_\text{eff} = k_1 + k_2\]

• This relationship implies that the combined system is stiffer than either spring alone.
• In the exercise, with \( k_1 = 1.8 \, \text{N/m} \) and \( k_2 = 3.2 \, \text{N/m} \), the effective spring constant is \( 5.0 \, \text{N/m} \).

This sum affects how the system will oscillate when the mass is introduced.

Oscillation Period Formula

The "Oscillation Period Formula" is critical in understanding how a mass attached to a spring behaves over time. The period of oscillation (\( T \)) indicates how long it takes for the mass to complete one full cycle of motion.For a mass-spring system, the formula to determine the period is given by:\[T = 2\pi\sqrt{\frac{m}{k_\text{eff}}}\]Where:

• \( m \) is the mass attached to the spring. In this exercise, it's \( 0.2\, \text{kg} \).
• \( k_\text{eff} \) is the effective spring constant. Here, it's \( 5.0\, \text{N/m} \).

Plugging these values into the formula allows us to compute the period, enabling the prediction of oscillation duration. After calculation, we find \( T = 1 + \frac{7\pi}{12} \). This exact time corresponds to one of the options provided in the exercise.

Spring System Dynamics

"Spring System Dynamics" involves understanding how the system of springs and the attached mass interact. This includes analyzing motion, forces, and resulting oscillations in the setup on a frictionless surface, such as the one described in the exercise.Key points to consider:

• The block will experience forces due to the compression or extension of each spring. This results in simple harmonic motion, characterized by regular oscillations around an equilibrium position.
• The mass dictates how fast it oscillates, influenced by the combined stiffness of the springs. The stronger the effective spring force, the quicker the potential oscillations.
• Being a frictionless surface ensures that energy losses are minimized, leading to consistent oscillations over time without damping.

These dynamics help predict future states of the system, vital in solving the exercise and selecting the accurate periodic time. The mass at \( 0.2\, \text{kg} \) and effective spring constant of \( 5.0\, \text{N/m} \) directly affect these dynamics, as outlined in the step-by-step calculation.