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Chapter 10: Problem 1386

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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Most popular questions from this chapter

Equation for a harmonic progressive wave is given by \(\mathrm{y}=\mathrm{A}\) \(\sin (15 \pi t+10 \pi x+\pi / 3)\) where \(x\) is in meter and \(t\) is in seconds. This wave is \(\ldots \ldots\) (A) Travelling along the positive \(\mathrm{x}\) direction with a speed of $1.5 \mathrm{~ms}^{-1}$ (B) Travelling along the negative \(\mathrm{x}\) direction with a speed of $1.5 \mathrm{~ms}^{-1} .$ (C) Has a wavelength of \(1.5 \mathrm{~m}\) along the \(-\mathrm{x}\) direction. (D) Has a wavelength of \(1.5 \mathrm{~m}\) along the positive \(\mathrm{x}\) - direction.

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