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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 cm and \(-\)15 cm/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
(a) The acceleration at \(t = 0\) is \(-48.7\) cm/s². (b) Position, velocity, and acceleration equations are: \(x(t)\), \(v(t)\), and \(a(t)\) respectively.

Step by step solution

01

Understand Simple Harmonic Motion

The point of the needle moves in simple harmonic motion (SHM), which follows the equation \(x(t) = A \cos(\omega t + \phi)\). The velocity is \(v(t) = -A \omega \sin(\omega t + \phi)\), and the acceleration is \(a(t) = -A \omega^2 \cos(\omega t + \phi)\).
02

Calculate Angular Frequency

Frequency \(f\) is given as 2.5 Hz. Angular frequency \(\omega\) is calculated by \(\omega = 2 \pi f\). Substitute \(f = 2.5\) Hz to find \(\omega\).
03

Find Phase Constant \(\phi\) using Initial Conditions

At \(t = 0\), the position \(x(0) = +1.1\) cm, and the velocity \(v(0) = -15\) cm/s. Use these values in the equations to find the amplitude \(A\) and the phase \(\phi\) by solving the system of equations given by the expressions for \(x(0)\) and \(v(0)\).
04

Find Amplitude \(A\)

Use initial position \(x(0) = A \cos(\phi) = 1.1\) and solve for \(A\). Then substitute into velocity equation after finding \(\phi\) from the system of equations.
05

Calculate Initial Acceleration

Use the equation for acceleration \(a(t) = -A \omega^2 \cos(\omega t + \phi)\). Substitute \(t = 0\), \(A\), \(\omega\), and \(\phi\) to find the acceleration at \(t = 0\).
06

Derive General Equations

Use the derived values from above steps to write down the general equations for position \(x(t)\), velocity \(v(t)\), and acceleration \(a(t)\) as functions of time.

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Key Concepts

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

Angular Frequency
Angular frequency is a critical concept in simple harmonic motion, representing how fast an object oscillates. It is denoted by the symbol \(\omega\) and is related to the frequency \(f\) of the oscillation. The relationship is given by the equation:\[\omega = 2 \pi f\]This means the angular frequency is what you'd get if you translated cycles per second (which is what frequency \(f\) measures) into radians per second, since a full cycle of oscillation is \(2 \pi\) radians.
For the sewing machine problem, with a frequency of 2.5 Hz, the angular frequency would be calculated by substituting into the equation, yielding \(\omega = 5\pi\). This tells us how quickly the point of the needle is oscillating back and forth in terms of radians per second.
Amplitude
Amplitude, symbolized by \(A\), measures the maximum extent of the oscillation from its equilibrium position in simple harmonic motion. It reflects how far the needle reaches on either side of the axis. In the context of our exercise, finding the amplitude involves knowing both the initial position and velocity.
Initially, using the equation for position:\[x(0) = A \cos(\phi) = 1.1\]The amplitude can be derived once the phase constant \(\phi\) is known. Amplitude reflects the energy of the motion; larger amplitudes mean the point moves further from the center, which usually requires more initial energy being imparted to the system.
Phase Constant
The phase constant \(\phi\) in simple harmonic motion provides information on where the oscillation begins at \(t = 0\). The phase constant essentially shifts the entire wave function along the time axis and can be critical in determining initial conditions.
To find \(\phi\) in our task, use both the given position and velocity at \(t=0\). With:\[x(0) = A \cos(\phi) = 1.1\]and\[v(0) = -A \omega \sin(\phi) = -15\]By solving these simultaneous equations, you determine \(\phi\). This accounts for the starting point of the needle's oscillation, capturing any phase shifts it might have.
Velocity and Acceleration Equations
Velocity and acceleration are dynamic components of simple harmonic motion that describe how quickly the object's position changes over time. They’re derived from the basic position function, and each adds dimensions to understanding the motion.The basic equations are:
  • Velocity: \(v(t) = -A \omega \sin(\omega t + \phi)\)
  • Acceleration: \(a(t) = -A \omega^2 \cos(\omega t + \phi)\)
The velocity is the derivative of position with respect to time, expressing how fast and in what direction the needle moves. It's directly influenced by both the amplitude \(A\) and angular frequency \(\omega\), explaining how fast and in what direction the needle is instantaneously moving.
The acceleration, being the derivative of velocity, provides insight into the forces in play to change that velocity. It incorporates the square of angular frequency, emphasizing how fast the oscillations increasingly or decreasingly act upon the needle. Understanding these relationships helps in predicting the needle's behavior over time, allowing for precise control in practical applications like sewing.

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Most popular questions from this chapter

An 85.0-kg mountain climber plans to swing down, starting from rest, from a ledge using a light rope 6.50 m long. He holds one end of the rope, and the other end is tied higher up on a rock face. Since the ledge is not very far from the rock face, the rope makes a small angle with the vertical. At the lowest point of his swing, he plans to let go and drop a short distance to the ground. (a) How long after he begins his swing will the climber first reach his lowest point? (b) If he missed the first chance to drop off, how long after first beginning his swing will the climber reach his lowest point for the second time?

A small block is attached to an ideal spring and is moving in SHM on a horizontal frictionless surface. The amplitude of the motion is 0.165 m. The maximum speed of the block is 3.90 m/s. What is the maximum magnitude of the acceleration of the block?

The tip of a tuning fork goes through 440 complete vibrations in 0.500 s. Find the angular frequency and the period of the motion.

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An object with mass 0.200 kg is acted on by an elastic restoring force with force constant 10.0 N/m. (a) Graph elastic potential energy \(U\) as a function of displacement \(x\) over a range of \(x\) from \(-\)0.300 m to \(+\)0.300 m. On your graph, let 1 cm \(=\) 0.05 J vertically and 1 cm \(=\) 0.05 m horizontally. The object is set into oscillation with an initial potential energy of 0.140 J and an initial kinetic energy of 0.060 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?

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