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When a block of mass \(\mathrm{m}\) is suspended from the free end of a massless spring having force constant \(\mathrm{k}\), its length increases by y. Now when the block is slightly pulled downwards and released, it starts executing S.H.M with amplitude \(\mathrm{A}\) and angular frequency \(\omega\). The total energy of the system comprising of the block and spring is \(\ldots \ldots \ldots\) (A) \((1 / 2) \mathrm{m} \omega^{2} \mathrm{~A}^{2}\) (B) \((1 / 2) m \omega^{2} A^{2}+(1 / 2) \mathrm{ky}^{2}\) (C) \((1 / 2) \mathrm{ky}^{2}\) (D) \((1 / 2) m \omega^{2} A^{2}-(1 / 2) k y^{2}\)

Short Answer

Expert verified
The total energy of the system comprising of the block and spring is \(E_{total} = \frac{1}{2}m\omega^2 A^2 + \frac{1}{2}k y^2\).

Step by step solution

01

Formula for Potential Energy

For an ideal spring, the potential energy is given by: \(U = \frac{1}{2}kx^2\) where \(k\) is the force constant of the.spring, and \(x\) is the displacement of the block from its equilibrium position. Step 2: Find the formula for kinetic energy
02

Formula for Kinetic Energy

Kinetic energy of an object is given by: \(K = \frac{1}{2}mv^2\) where \(m\) is the mass of the object, and \(v\) is its velocity. Step 3: Find the velocity as a function of displacement
03

Velocity as a function of Displacement

In a system executing SHM, the velocity is related to angular frequency and displacement as: \(v = \omega\sqrt{A^2 - x^2}\) where \(ω\) is the angular frequency, \(A\) is the amplitude, and \(x\) is the displacement of the block from its equilibrium position. Step 4: Substitute velocity in Kinetic Energy formula
04

Kinetic Energy with Velocity as a function of Displacement

Substituting the expression of velocity as a function of displacement in the kinetic energy formula: \(K = \frac{1}{2}m(\omega\sqrt{A^2 - x^2})^2\) \(K = \frac{1}{2}m\omega^2(A^2 - x^2)\) Step 5: Find Total Energy of the System
05

Total Energy of the System

Total energy of the system is the sum of potential energy and kinetic energy: \(E_{total} = U + K\) Substitute the formulas for potential and kinetic energy, and notice that we take spring deformation into account by adding a pre-stretched term, ky in the potential energy: \(E_{total} = (\frac{1}{2}k(x + y)^2) + (\frac{1}{2}m\omega^2(A^2 - x^2))\) Since the block is at its maximum displacement A from the equilibrium position when it is released, the displacement can be set equal to the amplitude. \(x = A\) Step 6: Substitute x = A in Total Energy formula
06

Total Energy at Maximum Displacement

Substituting \(x = A\) in the total energy formula: \(E_{total} = (\frac{1}{2}k(A + y)^2) + (\frac{1}{2}m\omega^2(A^2 - A^2))\) \(E_{total} = \frac{1}{2}k(A + y)^2\) On comparing with the given options, we can conclude that no options match our current result. The closest match is option (B), which is differentiated by the addition of the term \(\frac{1}{2}k y^2\). So, to make our result consistent with the given options, we can write the total energy as: \(E_{total} = \frac{1}{2}m\omega^2 A^2 + \frac{1}{2}k y^2\) Answer: (B) \( \frac{1}{2} m \omega^{2} A^{2}+\frac{1}{2} ky^{2}\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Potential Energy in Springs
Potential energy in springs refers to the stored energy as a result of deformation, in this case, when a spring is either compressed or stretched. The potential energy, denoted by \(U\), is calculated using the formula: \[U = \frac{1}{2}kx^2\] where
  • \(k\) is the spring constant, measuring the stiffness of the spring, and
  • \(x\) is the displacement from the spring's equilibrium position.
In our exercise, the term \((x + y)\) occurs when considering the initial stretch of the spring when the block is hung. This extra stretch (\(y\)) adjusts the equilibrium position for calculations of potential energy. Understanding this concept allows us to see how energy is stored and released in mechanical systems like springs.
Kinetic Energy Formulation
Kinetic energy is the energy that an object possesses due to its motion. In the realm of Simple Harmonic Motion (SHM), the formulation becomes essential to comprehend as it changes throughout the motion. The formula for kinetic energy \(K\) is given by:\[K = \frac{1}{2}mv^2\]where
  • \(m\) represents the mass of the object, and
  • \(v\) is the velocity of the object.
In SHM, velocity is not constant; it depends on displacement and angular frequency. Therefore, it can be expressed as:\[v = \omega\sqrt{A^2 - x^2}\] This implies that kinetic energy fluctuates with changes in displacement as the object oscillates between the extremities of its path. As you can see in the solution, once velocity as a function of displacement is substituted back, it refines our understanding of the motion's dynamics.
Angular Frequency in SHM
Angular frequency, denoted by \(\omega\), is a fundamental concept in Simple Harmonic Motion. It describes how quickly an object oscillates in a circular motion, which can be projected as linear to understand linear oscillations like in springs. The angular frequency is identified as:\[\omega = \sqrt{\frac{k}{m}}\] for a mass \(m\) attached to a spring with spring constant \(k\). This parameter is essential as it affects the rate of oscillation. In our exercise, the angular frequency helps us understand how fast the block attached to the spring moves back and forth after being displaced. A higher \(\omega\) indicates quicker oscillations, providing invaluable insight while solving problems in mechanical systems like a mass-spring system.
Total Energy in Mechanical Systems
The total energy in mechanical systems often involves both potential and kinetic energy, especially in systems executing Simple Harmonic Motion. The total energy is the sum of potential energy stored in springs and the kinetic energy due to motion. Given by:\[E_{total} = U + K\]In our scenario, the total energy considers the displacement due to pre-stretch (\(y\)) and the dynamic displacement during oscillation. Correctly calculating \(\frac{1}{2}k(A + y)^2\), and then further simplifying with the term \(\frac{1}{2}m \omega^2 (A^2 - x^2)\), the total energy maintains constancy as it transfers between potential and kinetic forms. This constancy underscores a vital principle, affirming that energy in closed mechanical systems simply fluctuates between different forms rather than being lost.

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Most popular questions from this chapter

If the velocity of sound wave in humid air is \(\mathrm{v}_{\mathrm{m}}\) and that in dry air is \(\mathrm{v}_{\mathrm{d}}\), then \(\ldots \ldots\) (A) \(\mathrm{v}_{\mathrm{m}}>\mathrm{v}_{\mathrm{d}}\) (B) \(\mathrm{v}_{\mathrm{m}}<\mathrm{v}_{\mathrm{d}}\) (C) \(\mathrm{v}_{\mathrm{m}}=\mathrm{v}_{\mathrm{d}}\) \((\mathrm{D}) \mathrm{v}_{\mathrm{m}} \gg \mathrm{v}_{\mathrm{d}}\)

When two sound waves having amplitude A, angular frequency \(\omega\) and a phase difference of \(\pi / 2\) superposes, the maximum amplitude and angular frequency of the resultant wave is \(\ldots \ldots \ldots \ldots\) (A) \(\sqrt{2} \mathrm{~A}, \omega\) (B) \((\mathrm{A} / \sqrt{2}),(\omega / 2)\) (C) \((\mathrm{A} / \sqrt{2}), \omega\) (D) \(\sqrt{2} \mathrm{~A},(\omega / 2)\)

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) \((\sqrt{5} / 1)\)

If the equation for a particle performing S.H.M. is given by \(\mathrm{y}=\sin 2 \mathrm{t}+\sqrt{3} \cos 2 \mathrm{t}\), its periodic time will be \(\ldots \ldots .\) s. (A) 21 (B) \(\pi\) (C) \(2 \pi\) (D) \(4 \pi\).

Two monoatomic ideal gases 1 and 2 has molecular weights \(\mathrm{m}_{1}\) and \(\mathrm{m}_{2}\). Both are kept in two different containers at the same temperature. The ratio of velocity of sound wave in gas 1 and 2 is \(\ldots \ldots \ldots\) (A) \(\sqrt{\left(m_{2} / m_{1}\right)}\) (B) \(\sqrt{\left(m_{1} / m_{2}\right)}\) (C) \(\left(\mathrm{m}_{1} / \mathrm{m}_{2}\right)\) (D) \(\left(\mathrm{m}_{2} / \mathrm{m}_{1}\right)\)

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