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A guitar string vibrates at a frequency of 440 Hz. A point at its center moves in SHM with an amplitude of 3.0 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.

Short Answer

Expert verified
Position: \(x(t) = 3.0 \cos(880\pi t)\). Max velocity: \(2640\pi \text{ mm/s}\). Max acceleration: \(3.0 \times 774400\pi^2 \text{ mm/s}^2\). Max jerk: \(3.0 \times 680244480\pi^3 \text{ mm/s}^3\).

Step by step solution

01

Write position equation

First, recognize the simple harmonic motion (SHM) equation for the position of the center of the string. The general formula for SHM is given by \( x(t) = A \cos(\omega t + \phi) \), where \(A\) is the amplitude, \(\omega\) is the angular frequency, and \(\phi\) is the phase angle. Given the amplitude \(A = 3.0 \text{ mm}\) and a phase angle of zero, the equation simplifies to \( x(t) = 3.0 \cos(\omega t) \).
02

Calculate angular frequency

The angular frequency \(\omega\) is related to the frequency \(f\) by \(\omega = 2\pi f\). Given the frequency \(f = 440 \text{ Hz}\), calculate \(\omega\) as follows: \(\omega = 2\pi \times 440 = 880\pi \text{ rad/s}\).
03

Position function equation

Now substitute \(\omega\) into the position equation: \( x(t) = 3.0 \cos(880\pi t) \). This is the position as a function of time for the center of the string.
04

Max velocity and acceleration

For SHM, the maximum velocity \(v_{\text{max}}\) is given by \(v_{\text{max}} = A\omega\). \(v_{\text{max}} = 3.0 \times 880\pi = 2640\pi \text{ mm/s}\). The maximum acceleration \(a_{\text{max}}\) is given by \(a_{\text{max}} = A\omega^2\). \(a_{\text{max}} = 3.0 \times (880\pi)^2 = 3.0 \times 774400\pi^2 \text{ mm/s}^2\).
05

Determine jerk function

The jerk is the derivative of acceleration with respect to time. Start by finding the acceleration function: \( a(t) = -A\omega^2 \cos(\omega t)\). Then, the jerk function is \( j(t) = \frac{d}{dt}[-A\omega^2 \cos(\omega t)] = A\omega^3 \sin(\omega t)\).
06

Maximum jerk calculation

The maximum value of the jerk \(j_{\text{max}}\) occurs when \(\sin(\omega t) = \pm 1\). \(j_{\text{max}} = A\omega^3 = 3.0 \times (880\pi)^3 = 3.0 \times 680244480\pi^3 \text{ mm/s}^3\).

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Key Concepts

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

Angular Frequency
Angular frequency, often denoted by the symbol \( \omega \), is an essential concept in the study of Simple Harmonic Motion (SHM). It's related to how quickly an oscillating object cycles through its motion, similar to how regular frequency measures cycles per second. Angular frequency is given by the formula \( \omega = 2\pi f \), where \( f \) is the frequency in hertz (Hz). For example, in the case of a guitar string vibrating at 440 Hz, the angular frequency would be \( \omega = 2\pi \times 440 = 880\pi \) rad/s. This means the string completes \( 880\pi \) radians every second. Angular frequency helps us determine other important properties of SHM, such as velocity and acceleration. Using it, we craft precise equations to describe motion, such as \( x(t) = A \cos(\omega t + \phi) \), aiding in predictions of the oscillating system's behavior over time.

Understanding the concept of angular frequency not only allows us to write accurate position equations but also helps in visualizing the rapidity of oscillation in physical systems. Its unit, radians per second, captures the conversion of periodic motion into rotational parameters, a valued insight for scientists and engineers alike.
Maximum Velocity and Acceleration
In Simple Harmonic Motion, both velocity and acceleration oscillate just like position does, reaching their extremums at specific points in the cycle. The maximum velocity \( v_{\text{max}} \) in SHM is derived from the formula \( v_{\text{max}} = A\omega \), where \( A \) is the amplitude and \( \omega \) is the angular frequency. Applying this to our guitar string example with amplitude of 3.0 mm and \( \omega = 880\pi \) rad/s, the maximum velocity becomes \( 2640\pi \) mm/s. This indicates the highest speed the string’s center reaches while oscillating.

Similarly, the maximum acceleration \( a_{\text{max}} \) is given by \( a_{\text{max}} = A\omega^2 \). For the guitar string, plugging in \( A \) and \( \omega \) gives us \( 3.0 \times 774400\pi^2 \) mm/s². The acceleration is at its fastest when the string passes through its equilibrium position, driven by the restoring force attempting to bring the system back to equilibrium. These extreme values help understand the energy dynamics at play—at maximum velocity the kinetic energy is at its peak, whereas maximum acceleration aligns with peak restoring force.
Jerk in Physics
Jerk, while perhaps a lesser-known term compared to velocity or acceleration, is crucial in understanding dynamic systems in more detail. It represents the rate of change of acceleration over time, defined as the derivative of acceleration. In terms of mathematics for SHM, it is expressed as \( j(t) = \frac{d}{dt}[-A\omega^2 \cos(\omega t)] = A\omega^3 \sin(\omega t) \). Breaking this down for a system with an amplitude of 3.0 mm and an angular frequency \( \omega = 880\pi \) rad/s, the maximum jerk \( j_{\text{max}} \) arises when \( \sin(\omega t) = \pm 1 \).

The result of \( 3.0 \times 680244480\pi^3 \) mm/s³ illustrates the maximum rate of change of the acceleration as the string vibrates. Jerk becomes especially important in scenarios where sudden changes in acceleration can impact the mechanical integrity or user comfort—for instance, the starting and stopping motions in vehicles or amusement parks rides. Accurate jerk calculations help in designing systems optimized for both performance and safety.

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Most popular questions from this chapter

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 0.250 m and the period is 3.20 s. What are the speed and acceleration of the block when \(x =\) 0.160 m?

An object is undergoing SHM with period 0.300 s and amplitude 6.00 cm. At \(t =\) 0 the object is instantaneously at rest at \(x =\) 6.00 cm. Calculate the time it takes the object to go from \(x =\) 6.00 cm to \(x = -\)1.50 cm.

In a physics lab, you attach a 0.200-kg air-track glider to the end of an ideal spring of negligible mass and start it oscillating. The elapsed time from when the glider first moves through the equilibrium point to the second time it moves through that point is 2.60 s. Find the spring's force constant.

A 0.500-kg glider, attached to the end of an ideal spring with force constant \(k =\) 450 N/m, undergoes SHM with an amplitude of 0.040 m. Compute (a) the maximum speed of the glider; (b) the speed of the glider when it is at \(x = -\)0.015 m; (c) the magnitude of the maximum acceleration of the glider; (d) the acceleration of the glider at \(x = -\)0.015 m; (e) the total mechanical energy of the glider at any point in its motion.

An object with height \(h\), mass \(M\), and a uniform cross-sectional area \(A\) floats upright in a liquid with density \(\rho\). (a) Calculate the vertical distance from the surface of the liquid to the bottom of the floating object at equilibrium. (b) A downward force with magnitude \(F\) is applied to the top of the object. At the new equilibrium position, how much farther below the surface of the liquid is the bottom of the object than it was in part (a)? (Assume that some of the object remains above the surface of the liquid.) (c) Your result in part (b) shows that if the force is suddenly removed, the object will oscillate up and down in SHM. Calculate the period of this motion in terms of the density \(\rho\) of the liquid, the mass \(M\), and the cross- sectional area A of the object. You can ignore the damping due to fluid friction (see Section 14.7).

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