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Chapter 10: Problem 17

A particle of mass \(m\) is executing oscillation about the origin on the X-axis. Its potential energy is \(U(x)=k|x|^{3}\), where \(k\) is a positive constant. If the amplitude of oscillation is \(a\), then its time period \(T\) is : (a) proportional to \(\frac{1}{\sqrt{a}}\) (b) independent of \(a\) (c) proportional to \(\sqrt{a}\) (d) proportional to \(a^{3 / 2}\)

Short Answer

Expert verified
The time period T is proportional to \( a^{3/2} \).

Step by step solution

01

Understanding the Potential Energy Function

Given the potential energy function \( U(x) = k|x|^3 \), we identify it as non-linear because the power of \(x\) is more than 2. This indicates the force is non-hookean, which is F = -dU/dx = -3kx^2.
02

Relating Force to Oscillation

To find the time period, we need the relationship between the force and oscillation. The equation of motion \( F = ma \) becomes \( m \ddot{x} = -3kx^2 \). This is not a simple harmonic oscillator, indicating an \
03

Consider Energy Conservation

Since no non-conservative forces are involved, total energy is conserved. At maximum amplitude, all energy is potential: \( U(a) = \frac{1}{2} mv^2(0) + 0 \, (K.E.) \).
04

Dimensional Analysis

Perform dimensional analysis on the oscillation to derive T's proportionality. Given the expressions involve terms like \( k \), \( a \), and \( m \), considering their dimensions leads us to explore \( T = C \cdot a^{n} \). Often for non-linear potential, \( T \propto a^{3/2} \).

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Key Concepts

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

Potential Energy Function
The potential energy function is a powerful tool in understanding nonlinear oscillations. Given that the potential energy in our scenario is described by the function \( U(x) = k|x|^3 \), we recognize this as a form of energy that varies with respect to the position \( x \), involving a non-linear relationship due to the absolute cube of \( x \). In such cases:
  • Unlike simple harmonic oscillators where potential energy is typically quadratic in nature \( U(x) = \frac{1}{2}kx^2 \), the cubic term indicates a steeper or more complex response curve.
  • This suggests a force on the particle that is non-linear, calculated as \( F = - \frac{dU}{dx} = -3kx^2 \).
  • Such non-linearity implies that the system does not respond to perturbations in the simple, predictable manner that linear systems do.
Identifying and understanding the nature of the potential energy function helps us predict how energy transitions between kinetic and potential forms during oscillation.
Energy Conservation
Energy conservation is a fundamental principle in physics, stating that within a closed system, energy is neither created nor destroyed, only transformed. In the context of our oscillating particle:
  • At the point of maximum amplitude, all the energy is stored as potential, meaning \( U(a) = k|a|^3 \).
  • When the particle passes through the origin, potential energy is zero, and kinetic energy is maximized.
  • Conserving energy ensures that the sum of kinetic and potential energy remains constant throughout the oscillation.
This consistent exchange between potential and kinetic energy without external loss or gain illustrates the oscillatory motion, and helps us calculate important properties like amplitude and velocity at various points in its cycle.
Dimensional Analysis
Dimensional analysis is a critical method to derive relationships between physical quantities based on their dimensions. For the oscillating particle system:
  • We use dimensions of mass \( m \), length \( L \), and time \( T \) to express the influential factors.
  • The potential energy term \( U = k|x|^3 \) behaves dimensionally as \( [k] = [M][L][T^{-2}] \).
  • To understand the periodic time \( T \), we analyze the dimensions, knowing the resultant relationship has to account for the amplitude \( a \).
Often for non-linear potentials like ours, dimensional analysis supports finding relationships such as \( T \propto a^{3/2} \), signifying the period's dependency on the amplitude.
Equation of Motion
The equation of motion describes how a particle responds to forces acting on it, encapsulating Newton's laws in mathematical form. Here:
  • The expression \( F = ma \) transforms into \( m \ddot{x} = -3kx^2 \), indicating a differential equation that defines motion.
  • This equation is non-linear due to the term \( x^2 \), unlike the linear \( x \) term found in simple harmonic motion.
  • Such an equation typically doesn't have a straightforward solution, requiring specialized methods or approximations for solutions like perturbation techniques.
This non-linear equation governs and describes the pattern of oscillation and helps in determining periods and amplitudes of motion, essential for deeper understanding of the system's behavior.

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Most popular questions from this chapter

There are two pendulums of length \(l_{1}\) and \(l_{2}\) whcih start vibrating. At some instant, the both pendulum are in mean position in the same phase. After how many vibrations of shorter pendulum, the both pendulum will be in phase in the mean position ? \(\left[\left(l_{1}>l_{2}\right), l_{1}=121 \mathrm{~cm}, l_{2}=100 \mathrm{~cm}\right]\) (a) 11 (b) 10 (c) 9 (d) 8

A block is performing S.H.M. along a vertical line with amplitude of \(40 \mathrm{~cm}\) on a horizontal plank. The block just lose the contact with plank when plank is momentarily at rest. Then : (Take \(g=10 \mathrm{~m} / \mathrm{s}^{-}\) ) (a) the period of its oscillation is \(\frac{2 \pi}{5} \mathrm{sec}\) (b) the period of its oscillation is \(\frac{2 \pi}{6} \mathrm{sec}\) (c) the period of its oscillation is \(\frac{\pi}{5} \mathrm{sec}\) (d) none of the above

A ring of mass \(m\) and radius \(R\) is pivoted at a point \(O\) on its periphery. It is free to rotate about an axis perpendicular to its plane. What is the period of ring ? (a) \(T=2 \pi \sqrt{\left(\frac{R}{g}\right)}\) (b) \(T=2 \pi \sqrt{\left(\frac{2 R}{g}\right)}\) (c) \(T=\pi \sqrt{\left(\frac{2 R}{g}\right)}\) (d) \(T=2 \pi \sqrt{\left(\frac{3 R}{g}\right)}\)

An elastic ball of density \(d\) is released and it falls through a height \(h\) before striking the surface of liquid of density \(\rho(d<\rho)\). The motion of ball is : (a) periodic (b) S.H.M. (c) circular (d) parabolic

A clock pendulum is adjusted for giving correct time in Patna. This clock pendulum also gives correct time in : (a) Delhi (b) Kota (c) Hyderabad (d) none of these
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