/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 13 The point of the needle of a sew... [FREE SOLUTION] | 91Ó°ÊÓ

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The point of the needle of a sewing machine moves in SHM along the \(x\) -axis with a frequency of 2.5 Hz. At \(t=0\) its position and velocity components are \(+1.1 \mathrm{cm}\) and \(-15 \mathrm{cm} / \mathrm{s},\) respectively. (a) Find the acceleration component of the needle at \(t=0\) . (b) Write equations giving the position, velocity, and acceleration components of the point as a function of time.

Short Answer

Expert verified
The acceleration at \(t=0\) is approximately \[-273.9\text{ cm/s}^2.\]

Step by step solution

01

Understand Simple Harmonic Motion (SHM)

In SHM, the motion is sinusoidal in time. The position, velocity, and acceleration can be expressed as: \[x(t) = A \cos(\omega t + \phi)\]\[v(t) = -A\omega \sin(\omega t + \phi)\]\[a(t) = -A\omega^2 \cos(\omega t + \phi)\]where \(A\) is amplitude, \(\omega\) is angular frequency, and \(\phi\) is the phase.
02

Determine Angular Frequency

The frequency given is \(2.5\text{ Hz}\). The angular frequency \(\omega\) is calculated using \(\omega = 2\pi f\):\[\omega = 2\pi \times 2.5 = 5\pi \text{ rad/s}\]
03

Calculate Initial Conditions

At \(t=0\), position \(x(0) = 1.1 \text{ cm}\), velocity \(v(0) = -15 \text{ cm/s}\). Use these to find amplitude \(A\) and phase \(\phi\).
04

Find Phase \(\phi\) Using Position

From \(x(0) = A \cos \phi\) and given that \(x(0) = 1.1\text{ cm}\), let \(A = \text{Variable}\). We will first use velocity equation to find \(A\).
05

Use Velocity to Find Amplitude and Phase

From\[v(0) = -A\omega\sin(\phi) = -15\text{ cm/s}\]Plug in \(\omega = 5\pi\) and solve:\[-A \times 5\pi \sin(\phi) = -15\]Solve for \(A\sin(\phi)\):\[A \sin(\phi) = \frac{15}{5\pi}\]
06

Equate Position and Velocity

Express \(\cos(\phi)\) in terms of known quantities:\[A \cos(\phi) = 1.1\]Now solve for \(A\) and \(\phi\) by squaring and adding the equations:\[A^2(\cos^2(\phi) + \sin^2(\phi)) = 1.1^2 + \left(\frac{15}{5\pi}\right)^2\]
07

Determine Amplitude \(A\)

By simplifying the above:\[A^2 = 1.21 + \left(\frac{15}{5\pi}\right)^2\approx 1.21 + 0.227^2 = 1.21 + 0.0515\]\[A = \sqrt{1.2615} \approx 1.122 cm\]
08

Calculate Phase \(\phi\)

Using \(A\cos(\phi) = 1.1\text{ cm}\):\[\cos(\phi) = \frac{1.1}{1.122}\]\[\cos(\phi) \approx 0.980\]\(\phi \approx \arccos(0.980)\approx 0.2 ext{ radians}\)
09

Find Acceleration at t=0

Use acceleration formula:\[a(0) = -A\omega^2 \cos(\phi)\]\[a(0) = -(1.122) \times (5\pi)^2 \times 0.980\]\[a(0) \approx -273.9\text{ cm/s}^2\]
10

Write Position, Velocity, and Acceleration as Functions of Time

The position, velocity, and acceleration are:\[x(t) = 1.122\cos(5\pi t + 0.2)\]\[v(t) = -1.122 \times 5\pi \sin(5\pi t + 0.2)\]\[a(t) = -1.122 \times (5\pi)^2 \cos(5\pi t + 0.2)\]

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Key Concepts

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

Angular Frequency
When dealing with Simple Harmonic Motion (SHM), understanding angular frequency is crucial. This concept describes how quickly an object moves through its cycle in SHM. Essentially, it tells us how fast the oscillations occur.

Angular frequency is denoted by the symbol \( \omega \) and is related to the regular frequency \( f \), which is the number of cycles per second. The relationship is defined by the formula:
  • \( \omega = 2\pi f \)
In this case, since the frequency is given as 2.5 Hz, using the formula we calculate:
  • \( \omega = 2\pi \times 2.5 = 5\pi \) rad/s
You might visualize angular frequency like how fast the hands on a clock move. The faster they move, the larger the angular frequency. This aspect is fundamental in describing the motion dynamics of any object undergoing SHM.
Amplitude
Amplitude in Simple Harmonic Motion (SHM) expresses the maximum extent of displacement from the equilibrium position.

In simpler terms, amplitude is how far the object swings either side of its rest position. It's denoted by the letter \( A \) and is always a positive value because it represents a distance.

In our exercise, after solving the equations, we found that the amplitude \( A \) of the sewing needle's motion is approximately \( 1.122 \) cm. Let's explore how to determine this:
  • Using the position equation, \( x(0) = A \cos(\phi) \) and the velocity equation, \( v(0) = -A\omega \sin(\phi) \), we could solve for both \( A \) and the phase angle \( \phi \).
  • For both the position and the velocity equations, we derived expressions involving \( \sin(\phi) \) and \( \cos(\phi) \), which were squared and summed to attain \( A^2 \).
  • Simplifying these results yielded \( A = \sqrt{1.2615} \approx 1.122 \) cm.
Amplitude is an important concept as it helps define the scale of oscillation in SHM, affecting how large or small the motion seems.
Phase Angle
The phase angle, often denoted by \( \phi \), is a vital component in understanding SHM, as it details the specific starting point of motion within its cycle.

It can seem a little abstract, as it doesn’t directly describe a physical property, but rather indicates the initial angle at \( t=0 \). Essentially, it helps us know whether the motion starts at the maximum displacement, the midpoint, or somewhere in between.
  • In our exercise, we first computed \( \phi \) using the known position and amplitude: \( \phi \approx \arccos(0.980) \approx 0.2 \text{ radians} \).
The phase angle makes sure that the resulting motion equations capture the initial conditions accurately, such as the given starting position and velocity.

This concept not only pinpoints the initial conditions but also helps in predicting future positions, velocities, and accelerations of objects exhibiting SHM.

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Most popular questions from this chapter

CP On the planet Newtonia, a simple pendulum having a bob with mass 1.25 \(\mathrm{kg}\) and a length of 185.0 \(\mathrm{cm}\) takes 1.42 \(\mathrm{s}\) . when released from rest, to swing through an angle of \(12.5^{\circ}\) , where it again has zero speed. The circumference of Newtonia is measured to be \(51,400 \mathrm{km}\) . What is the mass of the planet Newtonia?

CP SHM in a Car Engine. The motion of the piston of an automobile engine is approximately simple harmonic. (a) If the stroke of an engine (twice the amplitude) is 0.100 \(\mathrm{m}\) and the engine runs at 4500 \(\mathrm{rev} / \mathrm{min}\) , compute the acceleration of the piston at the endpoint of its stroke. (b) If the piston has mass \(0.450 \mathrm{kg},\) what net force must be exerted on it at this point? (c) What are the speed and kinetic energy of the piston at the mid- point of its stroke? (d) What average power is required to accelerate the piston from rest to the speed found in part (c)? (e) If the engine runs at 7000 rev/min, what are the answers to parts (b), (c), and (d)?

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.

CALC A 2.00 -kg bucket containing 10.0 \(\mathrm{kg}\) of water is hanging from a vertical ideal spring of force constant 125 \(\mathrm{N} / \mathrm{m}\) and oscillating up and down with an amplitude of 3.00 \(\mathrm{cm} .\) Suddenly the bucket springs a leak in the bottom such that water drops out at a steady rate of 2.00 \(\mathrm{g} / \mathrm{s}\) . When the bucket is half full, find (a) the period of oscillation and (b) the rate at which the period is changing with respect to time. Is the period getting longer or shorter? (c) What is the shortest period this system can have?

CALC Jerk. A guitar string vibrates at a frequency of 440 Hz. A point at its center moves in SHM with an amplitude of 3.0 \(\mathrm{mm}\) and a phase angle of zero. (a) Write an equation for the position of the center of the string as a function of time. (b) What are the maximum values of the magnitudes of the velocity and acceleration of the center of the string? (c) The derivative of the acceleration with respect to time is a quantity called the jerk. Write an equation for the jerk of the center of the string as a function of time, and find the maximum value of the magnitude of the jerk.

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