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Chapter 5: Problem 9

The maximum displacement of a mass oscillating about its equilibrium position is \(0.2 \mathrm{m},\) and its maximum speed is \(1.2 \mathrm{m} / \mathrm{s}\). What is the period \(\tau\) of its oscillations?

Short Answer

Expert verified
The period \(\tau\) is approximately 1.047 seconds.

Step by step solution

01

Understand the problem

We are given the maximum displacement (amplitude) of an oscillating mass as 0.2 meters and its maximum speed as 1.2 meters per second. We need to find the period, which is the time taken for one complete cycle of oscillation.
02

Use the formula for maximum speed in simple harmonic motion

For a mass oscillating in simple harmonic motion, the maximum speed (\(v_{ ext{max}}\)) is given by the formula:\[v_{ ext{max}} = A \cdot \omega\]where \(A\) is the amplitude and \(\omega\) is the angular frequency.
03

Calculate angular frequency

We rearrange the formula to solve for the angular frequency \(\omega\):\[\omega = \frac{v_{ ext{max}}}{A}\]Substitute the given values:\[\omega = \frac{1.2}{0.2} = 6 \text{ rad/s}\]
04

Relate angular frequency to period

The period \(\tau\) of simple harmonic motion is related to the angular frequency \(\omega\) by the formula:\[\tau = \frac{2\pi}{\omega}\].
05

Calculate the period

Substitute the calculated angular frequency into the formula:\[\tau = \frac{2\pi}{6}\]Simplify the expression:\[\tau \approx \frac{2\pi}{6} = \frac{\pi}{3} \approx 1.047 \text{ seconds}\]

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Key Concepts

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

Angular Frequency
In simple harmonic motion, the angular frequency, denoted as \( \omega \), plays a crucial role. It's a measurement of how quickly an object oscillates back-and-forth. It tells us how many radians the object covers per unit of time. You can think of angular frequency as the speed of oscillation.

For any oscillating system, the simplest formula for angular frequency is \( \omega = \frac{v_{\text{max}}}{A} \), where \( v_{\text{max}} \) is the maximum speed, and \( A \) is the maximum displacement (amplitude). Here, the unit of angular frequency is radians per second \( \text{rad/s} \).
  • Each complete oscillation corresponds to an angle of \( 2\pi \) radians.
  • Using \( \omega = 6 \text{ rad/s} \) means the mass completes one full cycle every second as it travels through 6 radians.
Understanding \( \omega \) helps in determining other characteristics of the oscillating system, such as the period of oscillation.
Maximum Displacement
Maximum displacement, also referred to as amplitude, represents the furthest point an oscillating object will reach from its equilibrium position. It's the peak value measured in meters (m) in this case.

In the problem, the maximum displacement given is 0.2 meters. This amplitude tells us both the extent of the object's swing and gives us necessary data to calculate other factors.
  • Amplitude influences both the energy of the system and how it behaves over time.
  • Greater amplitude means a larger, wider swing in the object's path.
Amplitude is a fundamental aspect in any study of oscillations, as it gives insight into the range and energy involved in the motion.
Oscillations
Oscillations refer to any repetitive back-and-forth movement around a central point or equilibrium position. In essence, they are cycles that the object repeats continuously over time.

Simple harmonic motion, like that discussed here, is characterized by sinusoidal oscillations, where both displacement and velocity vary smoothly and predictably. Each cycle returns to the starting point, completing an oscillation.
  • Oscillations occur in systems like pendulums, springs, and waves.
  • Key parameters defining oscillations include amplitude, period, and angular frequency.
The study of oscillations allows us to understand various natural and technological systems, from musical instruments to building designs that withstand seismic activity.
Period of Oscillation
The period of oscillation \( \tau \) refers to the time it takes to complete one full cycle of oscillation. Essentially, it's how long the "round trip" takes in the oscillation path.

In simple harmonic motion, there's a direct relationship between the period and angular frequency, given by \( \tau = \frac{2\pi}{\omega} \). By calculating \( \tau \), you can predict the regularity of oscillations.
  • Here, a \( \omega \) of 6 rad/s was used to find \( \tau \approx 1.047 \text{ seconds} \).
  • The period helps to understand the timing and rhythm of the oscillating system.
Accurately determining the period enables deeper insights into system behavior, such as synchronization in mechanical clocks or communication systems.

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Most popular questions from this chapter

The potential energy of two atoms in a molecule can sometimes be approximated by the Morse function, $$U(r)=A\left[\left(e^{(R-r) / S}-1\right)^{2}-1\right]$$ where \(r\) is the distance between the two atoms and \(A, R,\) and \(S\) are positive constants with \(S \ll R .\) Sketch this function for \(0 < r < \infty\). Find the equilibrium separation \(r_{\mathrm{o}}\), at which \(U(r)\) is minimum. Now write \(r=r_{\mathrm{o}}+x\) so that \(x\) is the displacement from equilibrium, and show that, for small displacements, \(U\) has the approximate form \(U=\) const \(+\frac{1}{2} k x^{2} .\) That is, Hooke's law applies. What is the force constant \(k ?\)

Consider a simple harmonic oscillator with period \(\tau\). Let \(\langle f\rangle\) denote the average value of any variable \(f(t),\) averaged over one complete cycle: $$\langle f\rangle=\frac{1}{\tau} \int_{0}^{\tau} f(t) d t$$ Prove that \(\langle T\rangle=\langle U\rangle=\frac{1}{2} E\) where \(E\) is the total energy of the oscillator. [Hint: Start by proving the more general, and extremely useful, results that \(\left\langle\sin ^{2}(\omega t-\delta)\right\rangle=\left\langle\cos ^{2}(\omega t-\delta)\right\rangle=\frac{1}{2} .\) Explain why these two results are almost obvious, then prove them by using trig identities to rewrite \(\sin ^{2} \theta\) and \(\left.\cos ^{2} \theta \text { in terms of } \cos (2 \theta) .\right]\)

Another interpretation of the \(Q\) of a resonance comes from the following: Consider the motion of a driven damped oscillator after any transients have died out, and suppose that it is being driven close to resonance, so you can set \(\omega=\omega_{\mathrm{o}}\). (a) Show that the oscillator's total energy (kinetic plus potential) is \(E=\frac{1}{2} m \omega^{2} A^{2} .\) (b) Show that the energy \(\Delta E_{\text {dis dissipated during one cycle by the damping force }}\) \(F_{\text {dmp }}\) is \(2 \pi m \beta \omega A^{2} .\) (Remember that the rate at which a force does work is \(F v .\) ) (c) Hence show that \(Q\) is \(2 \pi\) times the ratio \(E / \Delta E_{\text {dis: }}\)

Suppose that you have found a particular solution \(x_{\mathrm{p}}(t)\) of the inhomogeneous equation (5.48) for a driven damped oscillator, so that \(D x_{\mathrm{p}}=f\) in the operator notation of \((5.49) .\) Suppose also that \(x(t)\) is any other solution, so that \(D x=f .\) Prove that the difference \(x-x_{\mathrm{p}}\) must satisfy the corresponding homogeneous equation, \(D\left(x-x_{\mathrm{p}}\right)=0 .\) This is an alternative proof that any solution \(x\) of the inhomogeneous equation can be written as the sum of your particular solution plus a homogeneous solution; that is, \(x=x_{\mathrm{p}}+x_{\mathrm{h}}\).

The potential energy of a one-dimensional mass \(m\) at a distance \(r\) from the origin is $$U(r)=U_{0}\left(\frac{r}{R}+\lambda^{2} \frac{R}{r}\right)$$ for \(0 < r < \infty,\) with \(U_{\mathrm{o}}, R,\) and \(\lambda\) all positive constants. Find the equilibrium position \(r_{\mathrm{o}} .\) Let \(x\) be the distance from equilibrium and show that, for small \(x\), the PE has the form \(U=\) const \(+\frac{1}{2} k x^{2}\). What is the angular frequency of small oscillations?
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