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The jerk is defined to be the time rate of change of the acceleration. (a) If the velocity of an object undergoing SHM is given by \(v_{x}=-\omega A \sin (\omega t),\) what is the equation for the \(x\) -component of the jerk as a function of time? (b) What is the value of \(x\) for the object when the \(x\) -component of the jerk has its largest positive value? (c) What is \(x\) when the \(x\) -component of the jerk is most negative? (d) When it is zero? (e) If \(v_{x}\) equals \(-0.040 \mathrm{~s}^{2}\) times the \(x\) -component of the jerk for all \(t,\) what is the period of the motion?

Short Answer

Expert verified
The equations for jerk and acceleration as function of time are \(j_{x}=\omega^{3} A \sin(\omega t)\) and \(a_{x}=-\omega^{2} A \cos(\omega t)\) respectively. The position 'x' when the jerk is maximum, minimum, and zero are \(x = -A\), \(x = A\), and \(x = 0\) respectively. The period of the motion is 10s.

Step by step solution

01

Deriving Acceleration

To derive the jerk, first we derive the equation for acceleration (a), the rate of change of velocity. The acceleration \(a_{x}\) is the derivative of the velocity \(v_{x}\) with respect to time: \(a_{x}= \frac{dv_{x}}{dt}.\) This derivation yields the result: \(a_{x}=-\omega^{2} A \cos(\omega t)\)
02

Deriving Jerk

Now, derive the acceleration \(a_{x}\) with respect to time to obtain the jerk \(j_{x}\). The jerk \(j_{x}\) is given by: \(j_{x}= \frac{da_{x}}{dt}\). This derivation yields the result: \(j_{x}=\omega^{3} A \sin(\omega t)\)
03

Maximum and Minimum Jerk

The maximum and the minimum values of \(j_{x}\) happen when \(\sin(\omega t) = 1\) and \(\sin(\omega t) = -1\) respectively. Solve these equations for time to get \(t_{max}= \frac{\pi}{2\omega} + n \pi /\omega\) and \(t_{min}= \frac{3\pi}{2\omega} + n \pi /\omega\), where \(n\) is an integer.
04

Position X with Maximum, Minimum and Zero Jerk

Compute 'x' at these times using the equation for simple harmonic motion \(x = A \cos(\omega t)\). From this, you can find that when the jerk is maximum, \(x = -A\), when the jerk is minimum, \(x = A\), and when the jerk is zero, \(x = 0\) as \(\cos(\omega t)\) equals zero when \(\omega t = n\pi\), \(n\) being an integer.
05

Calculating the Period

Now, use the given relation \(v_{x} = -0.04 j_{x}\) to find the period. Plugging the derived values of \(v_{x}\) and \(j_{x}\), you get \(-\omega A \sin(\omega t) = -0.04\omega^{3} A \sin(\omega t)\). Dividing by \(-\omega A \sin(\omega t)\) and solving, you find that the period \(T = 2\pi/\omega = 10\) s.

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Key Concepts

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

Jerk (Physics)
In physics, the concept of jerk refers to the time rate of change of acceleration. Jerk is essentially how quickly an object's acceleration is increasing or decreasing, making it the third derivative of its position with respect to time.
  • Acceleration is the first rate of change of velocity,
  • while jerk is the second rate of change of velocity, or the derivative of acceleration.
Jerk is symbolized as \( j_{x} \) in formulas. During Simple Harmonic Motion (SHM), the object's velocity \(v_{x} = -\omega A \sin(\omega t)\) will change over time, leading to varying acceleration and, subsequently, jerk.
By taking the derivative of acceleration \( a_{x} = -\omega^{2} A \cos(\omega t) \), the jerk \( j_{x} \) can be derived:
  • The jerk \( j_{x} = \omega^{3} A \sin(\omega t) \).
  • This equation tells us that the jerk has the same form as velocity but is amplified by an extra factor of \( \omega^{2} \).
This shows that changes in acceleration occur in synchrony with the object's periodic motion, reaching peaks when \( \sin(\omega t) \) is either +1 or -1, and being zero when \( \sin(\omega t) \) is zero.
Acceleration in SHM
In Simple Harmonic Motion, acceleration plays a pivotal role as it determines the returning force acting on the object. Acceleration, denoted as \( a_{x} \), can be derived from the velocity equation \( v_{x} = -\omega A \sin(\omega t) \). It is calculated as:
  • \( a_{x} = \frac{dv_{x}}{dt} \)
  • This derivation leads to \( a_{x} = -\omega^{2} A \cos(\omega t) \).
Acceleration in SHM is directed toward the equilibrium position and is proportional to the displacement from that position.
The equation shows the following characteristics:
  • As \( x \) reaches its maximum positive or negative value, the acceleration reaches its maximum magnitude and is directed oppositely to the displacement.
  • When the object passes through the equilibrium point, \( x = 0 \), acceleration is zero because the restoring force is zero.
These properties of acceleration contribute to the periodic nature of SHM, ensuring that the object oscillates back and forth.
Period of Motion
The period of motion in Simple Harmonic Motion (SHM) is the time it takes for an object to complete one full cycle of oscillation. It is closely related to the angular frequency \( \omega \). The relation between angular frequency and the period \( T \) is given by:
  • \( T = \frac{2\pi}{\omega} \)
From the exercise, where it is known that the velocity \( v_{x} = -0.04\, j_{x} \), we can substitute the values derived before to find the period.
Using the following deductions:
  • \(-\omega A \sin(\omega t) = -0.04\omega^{3} A \sin(\omega t)\)
  • The period \( T \) simplifies to \( 2\pi/\omega \).
  • The calculation leads to \( T = 10 \) seconds.
This period signifies that each complete oscillation of the object in SHM takes 10 seconds. This long period suggests a system with lower frequency and potentially larger amplitude or mass, demonstrating the cyclical and regular natural behavior of SHM.

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Most popular questions from this chapter

A \(40.0 \mathrm{~N}\) force stretches a vertical spring \(0.250 \mathrm{~m}\). (a) What mass must be suspended from the spring so that the system will oscillate with a period of \(1.00 \mathrm{~s} ?\) (b) If the amplitude of the motion is \(0.050 \mathrm{~m}\) and the period is that specified in part (a), where is the object and in what direction is it moving \(0.35 \mathrm{~s}\) after it has passed the equilibrium position, moving downward? (c) What force (magnitude and direction) does the spring exert on the object when it is \(0.030 \mathrm{~m}\) below the equilibrium position, moving upward?

When an object of unknown mass is attached to an ideal spring with force constant \(120 \mathrm{~N} / \mathrm{m},\) it is found to vibrate with a frequency of \(6.00 \mathrm{~Hz}\). Find (a) the period of the motion; (b) the angular frequency; (c) the mass of the object.

Object \(A\) has mass \(m_{A}\) and is in \(\mathrm{SHM}\) on the end of a spring with force constant \(k_{A} .\) Object \(B\) has mass \(m_{B}\) and is in \(\mathrm{SHM}\) on the end of a spring with force constant \(k_{B}\). The amplitude \(A_{A}\) for object \(A\) is twice the amplitude \(A_{B}\) for the motion of object \(B\). Also, \(m_{B}=4 m_{A}\) and \(k_{A}=9 k_{B}\). (a) What is the ratio of the maximum speeds of the two objects, \(v_{\max , A} / v_{\max , B} ?\) (b) What is the ratio of their maximum accelerations, \(a_{\max , A} / a_{\max , B} ?\)

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

A \(10.0 \mathrm{~kg}\) mass is traveling to the right with a speed of \(2.00 \mathrm{~m} / \mathrm{s}\) on a smooth horizontal surface when it collides with and sticks to a second \(10.0 \mathrm{~kg}\) mass that is initially at rest but is attached to one end of a light, horizontal spring with force constant \(170.0 \mathrm{~N} / \mathrm{m}\). The other end of the spring is fixed to a wall to the right of the second mass. (a) Find the frequency, amplitude, and period of the subsequent oscillations. (b) How long does it take the system to return the first time to the position it had immediately after the collision?

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