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\(\mathrm{A} 0.500\) -kg mass on a spring has velocity as a function of time given by \(v_{x}(t)=-(3.60 \mathrm{cm} / \mathrm{s}) \sin \left[\left(4.71 \mathrm{s}^{-1}\right) t-\pi / 2\right]\) What are (a) the period; (b) the amplitude; (c) the maximum acceleration of the mass; (d) the force constant of the spring?

Short Answer

Expert verified
(a) 1.334 s, (b) 0.764 cm, (c) 16.95 cm/s², (d) 11.10 N/m.

Step by step solution

01

Identify given function details

We have the velocity function \(v_{x}(t) = -(3.60 \ \mathrm{cm/s}) \sin\left((4.71 \ \mathrm{s^{-1}})t - \pi/2\right)\). This function is of the form \(v(t) = -v_{max} \sin(\omega t + \phi)\) where \(v_{max} = 3.60 \ \mathrm{cm/s}\), \(\omega = 4.71 \ \mathrm{s^{-1}}\), and \(\phi = -\pi/2\)."
02

Calculate the period

The angular frequency \(\omega\) is related to the period \(T\) by the equation \(\omega = \frac{2\pi}{T}\). So, \(T = \frac{2\pi}{\omega} = \frac{2\pi}{4.71} \approx 1.334 \ \mathrm{s}\).
03

Determine the amplitude

The maximum velocity \(v_{max} = \omega A\), where \(A\) is the amplitude. Therefore, \(A = \frac{v_{max}}{\omega} = \frac{3.60 \ \mathrm{cm/s}}{4.71 \ \mathrm{s^{-1}}} \approx 0.764 \ \mathrm{cm}\).
04

Find the maximum acceleration

Maximum acceleration \(a_{max}\) is given by \(a_{max} = \omega^2 A\). Substituting \(\omega = 4.71 \ \mathrm{s^{-1}}\) and \(A = 0.764 \ \mathrm{cm}\), we have \(a_{max} = (4.71)^2 \times 0.764 \ \mathrm{cm/s^2} \approx 16.95 \ \mathrm{cm/s^2}\).
05

Calculate the force constant of the spring

The force constant \(k\) is related to \(\omega\) and mass \(m\) by \(\omega = \sqrt{\frac{k}{m}}\). Thus, \(k = m\omega^2\). Substituting \(m = 0.500 \ \mathrm{kg}\) and \(\omega = 4.71 \ \mathrm{s^{-1}}\), we find \(k = 0.500 \times (4.71)^2 \approx 11.10 \ \mathrm{N/m}\).

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Key Concepts

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

Angular Frequency
Let's delve into angular frequency, symbolized by \( \omega \). It is a measure of how quickly an object undergoing simple harmonic motion oscillates. Angular frequency is expressed in radians per second - Imagine the swinging motion of a pendulum. Angular frequency tells us how fast it goes back and forth.- Learn how to calculate angular frequency using the period \( T \) (the time it takes for one full cycle):\[ \omega = \frac{2\pi}{T} \]- It's important to note that a higher angular frequency means a faster oscillation. In our given problem, we determined that \( \omega = 4.71 \ \mathrm{s^{-1}} \). This tells us that the oscillation completes quickly, in roughly 1.334 seconds, per full cycle.
Amplitude
Amplitude, denoted as \( A \), is the peak value of an oscillation. It's the furthest distance the object travels from its resting position.- In simple terms, think of amplitude as the height of a wave in analogy to the water waves - For mechanical systems like a mass attached to a spring, amplitude measures how far the mass can stretch from its equilibrium point.- For velocity in harmonic motion, the maximum velocity \( v_{max} \) relates to amplitude via \( v_{max} = \omega A \).From our problem, we calculated the amplitude as \( A = 0.764 \ \mathrm{cm} \). This small amplitude indicates a gentle motion rather than a vigorous one, emphasizing subtle oscillations.
Force Constant
Consider the force constant \( k \), which is a measure of how stiff or strong a spring is. In the realm of oscillations, \( k \) defines the restoring force's effectiveness in Newton's law for springs, expressed by Hooke's law: \( F = -kx \).- If the force constant is high, the spring is stiff and hard to stretch.- It is directly related to angular frequency and mass: \[ \omega = \sqrt{\frac{k}{m}} \]- The relationship shows that with a greater force constant, there is a greater angular frequency for the same mass.For the given problem, \( k \) was found to be approximately \( 11.10 \ \mathrm{N/m} \), indicating the spring provides a moderately strong restoring force against displacement.
Maximum Acceleration
Maximum acceleration represents how fast an object can accelerate as it moves through its oscillation.- It's the peak acceleration experienced by the mass when it is at its two endpoints (maximum displacement) during the motion.- In simple harmonic motion, acceleration is directly tied to both angular frequency and amplitude: \[ a_{max} = \omega^2 A \]- A greater angular frequency or amplitude means more energy to accelerate quickly, resulting in a larger maximum acceleration.For our oscillating mass, the maximum acceleration was calculated as approximately \( 16.95 \ \mathrm{cm/s^2} \), showing how swiftly it can move from the peak of its motion back to the equilibrium.

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Most popular questions from this chapter

A uniform beam is suspended horizontally by two identical vertical springs that are attached between the ceiling and each end of the beam. The beam has mass 225 \(\mathrm{kg}\) , and a 175 -kg sack of gravel sits on the middle of it. The beam is oscillating in \(\mathrm{SHM}\) , with an amplitude of 40.0 \(\mathrm{cm}\) and a frequency of 0.600 cycle/s. (a) The sack of gravel falls off the beam when the beam has its maximum upward displacement. What are the frequency and amplitude of the subsequent SHM of the beam? (b) If the gravel instead falls off when the beam has its maximum speed, what are the frequency and amplitude of the subsequent SHM of the beam?

An object with mass 0.200 \(\mathrm{kg}\) is acted on by an elastic restoring force with force constant 10.0 \(\mathrm{N} / \mathrm{m}\) . (a) Graph elastic potential energy \(U\) as a function of displacement \(x\) over a range of \(x\) from \(-0.300 \mathrm{m}\) to \(+0.300 \mathrm{m}\) . On your graph, let \(1 \mathrm{cm}=0.05 \mathrm{J}\) vertically and \(1 \mathrm{cm}=0.05 \mathrm{m}\) horizontally. The object is set into oscillation with an initial potential energy of 0.140 \(\mathrm{J}\) and an initial kinetic energy of 0.060 \(\mathrm{J}\) . Answer the following questions by referring to the graph. (b) What is the amplitude of oscillation? (c) What is the potential energy when the displacement is one half the amplitude? (d) At what displacement are the kinetic and potential energies equal? (e) What is the value of the phase angle \(\phi\) if the initial velocity is positive and the initial displacement is negative?

A tuning fork labeled 392 Hz has the tip of each of its two prongs vibrating with an amplitude of 0.600 \(\mathrm{mm} .\) (a) What is the maximum speed of the tip of a prong? (b) A housefly (Musca domestica) with mass 0.0270 \(\mathrm{g}\) is holding onto the tip of one of the prongs. As the prong vibrates, what is the fly's maximum kinetic energy? Assume that the fly's mass has a negligible effect on the frequency of oscillation.

A 2.50 -kg rock is attached at the end of a thin, very light rope 1.45 \(\mathrm{m}\) long. You start it swinging by releasing it when the rope makes an \(11^{\circ}\) angle with the vertical. You record the observation that it rises only to an angle of \(4.5^{\circ}\) with the vertical after 10\(\frac{1}{2}\) swings. (a) How much energy has this system lost during that time? (b) What happened to the "lost" energy? Explain how it could have been "lost."

A proud deep-sea fisherman hangs a 65.0 -kg fish from an ideal spring having negligible mass. The fish stretches the spring 0.120 \(\mathrm{m}\) . (a) Find the force constant of the spring. The fish is now pulled down 5.00 \(\mathrm{cm}\) and released. (b) What is the period of oscillation of the fish? (c) What is the maximum speed it will reach?

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