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Chapter 17: Problem 20

(a) When the displacement is one-half the amplitude \(x_{m}\), what fraction of the total energy is kinetic and what fraction is potential in simple harmonic motion? (b) At what displacement is the energy half kinetic and half potential?

Short Answer

Expert verified
For(a), when \(x = 0.5 x_m\), the kinetic energy is \(K/E = 0.75\) and the potential energy is \(U/E = 0.25\). For(b), the energy is half kinetic and half potential when the displacement is \(x = x_m / sqrt{2}\)

Step by step solution

01

Understand Energy Distribution in Simple Harmonic Motion

In simple harmonic motion, energy continually transfers back and forth between kinetic and potential energy, however the total energy is conserved. At maximum displacement, i.e. at amplitude \(x_m\), all energy is potential, and at equilibrium (where displacement is zero), all energy is kinetic. The kinetic energy K and the potential energy U at any point can be described using these formulas:\(K = 0.5 m v^2\), \(U = 0.5 k x^2\), where m denotes the mass, v the speed, k the spring constant and x the displacement
02

Find the Kinetic and Potential Energy Fraction with Displacement \(0.5 x_m\)

When the displacement is half the amplitude (\(x = 0.5 x_m\)), using formulas for kinetic and potential energy, the total energy E is \(E = K + U = 0.5 k x^2 + 0.5 m v^2\). But the total energy is also equal to the potential energy at maximum displacement (\(x = x_m\)), so \(E = 0.5 k x_m^2\). Therefore, the fractions of kinetic and potential energy are \(K/E = 0.5 m v^2 / 0.5 k x_m^2\) and \(U/E = 0.5 k x^2 / 0.5 k x_m^2\), respectively. Note here that \(v^2 = ω^2 (x_m^2 - x^2)\), where ω is the angular frequency.
03

Determine the Displacement for Half Kinetic and Half Potential Energy

When the energy is half kinetic and half potential, it implies \(K = U\). Using the formulas \(K = 0.5 m v^2\) and \(U = 0.5 k x^2\), you find \(0.5 m v^2 = 0.5 k x^2\). Remember \(v = ω sqrt{x_m^2 - x^2}\), then solve for x. This gives the result \(x = x_m / sqrt{2}\).

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Most popular questions from this chapter

Starting from Eq. \(17-43\), find the velocity \(v_{x}(=d x / d t)\) in forced oscillatory motion. Show that the velocity amplitude is $$ v_{\mathrm{m}}=F_{\mathrm{m}} /\left[\left(m \omega^{\prime \prime}-k / \omega^{\prime}\right)^{2}+b^{2}\right]^{1 / 2} $$ The equations of Section \(17-8\) are identical in form with those representing an electrical circuit containing a resistance \(R\), and inductance \(L\), and a capacitance \(C\) in series with an alternating emf \(V=V_{\mathrm{m}} \cos \omega^{\prime \prime} t\). Hence \(b, m, k\), and \(F_{\mathrm{m}}\), are analogous to \(R, L, 1 / C\), and \(V_{\mathrm{m}}\), respectively, and \(x\) and \(v\) are analogous to electric charge \(q\) and current \(i\), respectively. In the electrical case the current amplitude \(i_{\mathrm{m}}\), analogous to the velocity amplitude \(v_{\mathrm{m}}\) above, is used to describe the quality of the resonance.

An oscillator consists of a block attached to a spring \((k=\) \(456 \mathrm{~N} / \mathrm{m}) .\) At some time \(t\), the position (measured from the equilibrium location), velocity, and acceleration of the block are \(x=0.112 \mathrm{~m}, v_{x}=-13.6 \mathrm{~m} / \mathrm{s}, a_{x}=-123 \mathrm{~m} / \mathrm{s}^{2} .\) Calcu- late \((a)\) the frequency, \((b)\) the mass of the block, and \((c)\) the amplitude of oscillation.

An oscillating block-spring system has a mechanical energy of \(1.18 \mathrm{~J}\), an amplitude of \(9.84 \mathrm{~cm}\), and a maximum speed of \(1.22 \mathrm{~m} / \mathrm{s} .\) Find \((a)\) the force constant of the spring, \((b)\) the mass of the block, and \((c)\) the frequency of oscillation.

A \(5.22-\mathrm{kg}\) object is attached to the bottom of a vertical spring and set vibrating. The maximum speed of the object is \(15.3 \mathrm{~cm} / \mathrm{s}\) and the period is \(645 \mathrm{~ms}\). Find \((a)\) the force constant of the spring, (b) the amplitude of the motion, and \((c)\) the frequency of oscillation.

A damped harmonic oscillator involves a block (m = \(1.91 \mathrm{~kg}\) ), a spring \((k=12.6 \mathrm{~N} / \mathrm{m})\), and a damping force \(F=\) \(-b v_{x} .\) Initially, it oscillates with an amplitude of \(26.2 \mathrm{~cm}\); because of the damping, the amplitude falls to three-fourths of this initial value after four complete cycles. (a) What is the value of \(b ?(b)\) How much energy has been "lost" during these four cycles?
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