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A particle vibrates according to the equation \(x=20 \cos 16 t\), where \(x\) is in centimeters. Find its amplitude, frequency, and position at exactly \(t=0 \mathrm{~s}\).

Short Answer

Expert verified
The amplitude is 20 cm, frequency is approximately 2.55 Hz, and position at \(t=0\) is 20 cm.

Step by step solution

01

Identify the Amplitude

The amplitude of the motion in a cosine function described as \(x = A \cos(\omega t)\) is the coefficient \(A\) in front of the cosine function, which represents the maximum displacement from the mean position. In the given equation \(x = 20 \cos(16t)\), the amplitude is \(20\) cm.
02

Determine the Angular Frequency

The term inside the cosine function, \(\omega t\), represents the angular frequency \(\omega\). From the equation \(x = 20 \cos(16t)\), we can identify \(\omega = 16\) rad/s.
03

Convert Angular Frequency to Frequency

The frequency \(f\) of the motion is related to the angular frequency \(\omega\) by the formula \(f = \frac{\omega}{2\pi}\). Substitute \(\omega = 16\) rad/s into the formula: \(f = \frac{16}{2\pi} \approx 2.55\) Hz.
04

Find the Position at t = 0

Substitute \(t = 0\) into the equation \(x = 20 \cos(16t)\). At \(t = 0\), the expression becomes \(x = 20 \cos(0)\). Since \(\cos(0) = 1\), the position \(x = 20 \times 1 = 20\) cm.

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Key Concepts

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

Amplitude
Amplitude is a fundamental concept in harmonic motion. It refers to the maximum extent of displacement from the mean or equilibrium position that a particle achieves in its oscillation. When we look at the standard equation of motion for harmonic oscillators like \( x = A \cos(\omega t) \), the amplitude is represented by \( A \). In simpler terms, it's the "height" of the wave if you imagine the oscillation on a graph.

For the equation given in the problem \( x = 20 \cos(16t) \), the amplitude is clearly \( 20 \) cm. This means that the particle moves 20 cm away from its central position during its maximum swing. Understanding amplitude is crucial because it directly affects the energy of the system:
  • More amplitude, more potential energy at the extremities.
  • Determines the "loudness" if it's a sound wave analogy.
  • Affects the initial energy levels in mechanical movements.
Amplitude doesn't change as the motion continues unless an external force acts upon it, like damping or input energy.
Angular Frequency
Angular frequency is a measure of how quickly an object is moving through its cycle in harmonic motion. It's usually denoted by \( \omega \) and is expressed in radians per second. In essence, it quantifies how many radians the wave covers per unit time.

In the equation \( x = 20 \cos(16t) \), the term \( 16 \) appears alongside the trigonometric function, representing \( \omega \). This signifies that the angular frequency is \( 16 \) rad/s. Understanding angular frequency helps us connect to circular motion concepts:
  • Larger \( \omega \) means faster motion through cycles.
  • It links linear oscillations to rotational ones.
  • In systems like electrical engineering, it defines signal bandwidth.
This parameter is vital for interpreting how different systems, from pendulums to springs, react over time.
Frequency
Frequency is another vital concept closely associated with harmonic motion. Represented typically by \( f \), it indicates how many complete cycles or oscillations occur in a unit of time, measured in Hertz (Hz). While angular frequency relates to radians, frequency gives us a more intuitive sense of cycles per second.

To convert angular frequency \( \omega \) to frequency \( f \), we use the formula:
\[ f = \frac{\omega}{2\pi}\]

Applying this to our angular frequency from earlier (\( \omega = 16 \) rad/s), we get a frequency of \( f \approx 2.55 \) Hz. In practical terms:
  • Higher frequencies mean quicker oscillations.
  • Essential for understanding wave transmission and reception.
  • Key factor in designing resonant systems.
Frequency helps bridge everyday phenomena like sound and light to the complexities of mathematical physics.
Particle Position
Particle position in harmonic motion describes the exact location of a particle at any given time. This is pivotal to understanding how particles move back and forth in systems described by sinusoidal equations.

In the exercise problem \( x = 20 \cos(16t) \), we sought the particle's position at \( t = 0 \) s. By substituting \( t = 0 \) into the equation, we evaluate:\[ x = 20 \cos(16 \times 0) = 20 \cos(0) = 20 \times 1 = 20 \text{ cm} \]This tells us that the particle is at its maximum positive amplitude when the time is zero.

Particle position is dynamic and changes over time as \( t \) increases or decreases. Key bits to remember include:
  • Determining position involves time substitution.
  • Position oscillates symmetrically around an equilibrium point.
  • Understanding position helps visualize physical oscillations, such as springs or pendulums.
Grasping particle position in varying scenarios allows for deeper insights into how harmonic motion operates in real-life systems.

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Most popular questions from this chapter

In an old gasoline engine, a piston undergoes vertical SHM with an amplitude of \(7.0 \mathrm{~cm}\). A washer rests on top of the piston. As the motor speed is slowly increased, at what frequency will the washer no longer stay in contact with the piston? The situation we are looking for is when the maximum downward acceleration of the washer equals that of free fall, namely, \(g\). If the piston accelerates down faster than that, the washer will lose contact. In SHM, the acceleration is given in terms of the displacement and the period as $$ a=-\frac{4 \pi^{2}}{T^{2}} x $$ (To see this, notice that \(a=-\mathrm{F} / \mathrm{m}\). But from \(T=2 \pi \sqrt{\mathrm{m} / k}\), we have \(k=4 \pi^{2} m / T^{2}\), which then gives the above expression for \(a\).) With the upward direction chosen as positive, the largest downward (most negative) acceleration occurs for \(x=+x_{0}=0.070 \mathrm{~m}\); it is $$ a_{0}=\frac{4 \pi^{2}}{T^{2}}(0.070 \mathrm{~m}) $$ The washer will separate from the piston when \(a_{0}\) first becomes equal to \(g\). Therefore, the critical period for the SHM, \(T_{c}\), is given by $$ \frac{4 \pi^{2}}{T_{c}^{2}}(0.070 \mathrm{~m})=g \quad \text { or } \quad T_{c}=2 \pi \sqrt{\frac{0.070 \mathrm{~m}}{g}}=0.53 \mathrm{~s} $$ This corresponds to the frequency \(f_{c}=1 / T_{c}=1.9 \mathrm{~Hz}\). The washer will separate from the piston if the piston's frequency exceeds \(1.9\) cycles/s.

A "seconds pendulum" beats seconds; that is, it takes \(1 \mathrm{~s}\) for half a cycle. (a) What is the length of a simple "seconds pendulum" at a place where \(g=9.80 \mathrm{~m} / \mathrm{s}^{2} ?(b)\) What is the length there of \(a\) pendulum for which \(T=1.00 \mathrm{~s}\) ?

A pendulum is timed as it swings back and forth. The clock is started when the bob is at the left end of its swing. When the bob returns to the left end for the 90 th return, the clock reads \(60.0 \mathrm{~s}\). What is the period of vibration? The frequency?

A 300 -g object attached to the end of a spring oscillates with an amplitude of \(7.0 \mathrm{~cm}\) and a frequency of \(1.80 \mathrm{~Hz}\). (a) Find its maximum speed and maximum acceleration. (b) What is its speed when it is \(3.0 \mathrm{~cm}\) from its equilibrium position?

A 50 -g mass vibrates in SHM at the end of a spring. The amplitude of the motion is \(12 \mathrm{~cm}\), and the period is \(1.70 \mathrm{~s}\). Find: (a) the frequency, (b) the spring constant, (c) the maximum speed of the mass, \((d)\) the maximum acceleration of the mass, \((e)\) the speed when the displacement is \(6.0 \mathrm{~cm}\) and the mass is moving to the right, and \((f)\) the acceleration when \(x=6.0 \mathrm{~cm}\) and the mass is moving to the right. (a) \(f=\frac{1}{T}=\frac{1}{1.70 \mathrm{~s}}=0.588 \mathrm{H}_{2}\) (o) Sibce \(T=2 \pi \sqrt{m / k}\) $$ k=\frac{4 \pi^{2} m}{T^{2}}=\frac{4 \pi^{2}(0.050 \mathrm{~kg})}{(1.70 \mathrm{~s})^{2}}=0.68 \mathrm{~N} / \mathrm{m} $$ (c) \(\left|u_{0}\right|=x_{0} \sqrt{\frac{k}{m}}=(0.12 \mathrm{~m}) \sqrt{\frac{0.68 \mathrm{~N} / \mathrm{m}}{0.050 \mathrm{~kg}}}=0.44 \mathrm{~m} / \mathrm{s}\) (d) From \(a=-(k / m) x\) it is seen that \(a\) has maximum magnitude when \(x\) has maximum magnitude, that is, at the endpoints \(x=\) \(\pm x_{0}\). Thus the magnitude of the maximum acceleration \(\left(a_{0}\right.\) or \(a_{\max }\) ) is given by $$ a_{0}=\frac{k}{m} x_{0}=\frac{0.68 \mathrm{~N} / \mathrm{m}}{0.050 \mathrm{~kg}}(0.12 \mathrm{~m})=1.6 \mathrm{~m} / \mathrm{s}^{2} $$ where magnitudes are always positive. (e) From \(|v|=\sqrt{\left(x^{2}-x^{2}\right)(k / m)}\) $$ |v|=\sqrt{\frac{\left[(0.12 \mathrm{~m})^{2}-(0.06 \mathrm{~m})^{2}\right](0.6 \mathrm{~N} / \mathrm{m})}{(0.050 \mathrm{~kg})}}=0.38 \mathrm{~m} / \mathrm{s} $$ and, as ever, speed is positive. (f) Here we want the acceleration. Since \(x=6.0 \mathrm{~cm}\), the force on the mass is to the left and negative. Likewise the mass is accelerating to the left even as it is moving to the right. Hence the acceleration must be negative; the mass is slowing down. $$ a=-\frac{k}{m} x=-\frac{0.68 \mathrm{~N} / \mathrm{m}}{0.050 \mathrm{~kg}}(0.060 \mathrm{~m})=-0.82 \mathrm{~m} / \mathrm{s}^{2} $$

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