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Chapter 11: Problem 35

A particle that is at the origin of coordinates at exactly \(t=0\) vibrates about the origin along the \(y\) -axis with a frequency of \(20 \mathrm{~Hz}\) and an amplitude of \(3.0 \mathrm{~cm}\). Write out its equation of motion in centimeters.

Short Answer

Expert verified
The equation of motion is \( y(t) = 3.0 \sin(40\pi t) \) in centimeters.

Step by step solution

01

Understand the Equation of Motion

The equation of motion for a vibrating particle is described by the harmonic motion equation: \[ y(t) = A \sin(2\pi f t + \phi) \]where \(A\) is the amplitude, \(f\) is the frequency, and \(\phi\) is the phase shift. Since the particle is at the origin at \(t=0\), the phase \(\phi\) is 0.
02

Assign Known Values

Now we assign the known values to the equation:- Amplitude, \(A = 3.0 \text{ cm}\)- Frequency, \(f = 20 \text{ Hz}\)- Phase shift, \(\phi = 0\)The equation simplifies to:\[ y(t) = 3.0 \sin(2\pi \times 20 \times t) \]
03

Simplify the Equation

Substitute \(2\pi \times 20\) in the equation to simplify it further. This results in:\[ y(t) = 3.0 \sin(40\pi t) \]
04

Write the Final Equation of Motion

The equation of motion for the particle in centimeters is:\[ y(t) = 3.0 \sin(40\pi t) \]

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Key Concepts

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

Equation of Motion
Motion equations describe how objects move relative to time. In the context of harmonic motion, the equation of motion has a standard form: \[ y(t) = A \sin(2\pi f t + \phi) \]. Here are the meanings of all the symbols:
  • \( y(t) \) is the particle's displacement from a reference point (like an origin) at time \( t \).
  • \( A \) represents amplitude, showing the maximum displacement.
  • \( f \) is the frequency of the oscillation.
  • \( \phi \) is the phase shift, which affects the starting position of the motion.
The significance of the equation is that it represents periodic motion, common in wave and sound physics. In essence, the equation tells you where the particle is, given a specific time. This makes it a handy tool for analyzing movements like oscillations.
Frequency
Frequency in harmonic motion tells how many cycles or vibrations occur in one second. It is measured in Hertz (Hz). In our problem, the frequency is given as 20 Hz, which means the particle completes 20 full vibrations or oscillations every second. Frequency is crucial because:
  • It influences the speed of the vibration.
  • Higher frequencies mean more rapid vibrations.
  • It helps in classifying the type of wave motion, such as audible sound.
Understanding frequency helps you determine the dynamic behavior of the particle in motion. It reflects how quickly an oscillating system repeats its motion.
Amplitude
Amplitude in the context of harmonic motion measures the maximum distance from the point of equilibrium (origin) the particle reaches during its vibration. In our specific exercise, the amplitude is 3.0 cm. Amplitude helps us understand the extent of oscillation. Key insights about amplitude:
  • It provides a measure of how strong or intense the vibration is.
  • Larger amplitudes indicate larger displacements.
  • In physical systems, amplitude might correlate with energy in the motion.
Knowing the amplitude helps you visualize the full range the particle will move as it oscillates, making it easier to predict its behavior in motion.

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Most popular questions from this chapter

When a 400 -g mass is hung at the end of a vertical spring, the spring stretches \(35 \mathrm{~cm}\). Determine the elastic constant of the spring. How much farther will it stretch if an additional \(400-\mathrm{g}\) mass is hung from it? Use \(F_{\text {eut }}=k y\), where that force is the weight of the hanging mass: $$ F_{\text {eut }}=m g=(0.400 \mathrm{~kg})\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)=3.92 \mathrm{~N} $$ Therefore, $$ k=\frac{F_{\text {ext }}}{y}=\frac{3.92 \mathrm{~N}}{0.35 \mathrm{~m}}=11.2 \mathrm{~N} / \mathrm{m} \quad \text { or } \quad 11 \mathrm{~N} / \mathrm{m} $$ Once the elastic constant is known, we can determine how the spring will behave. With an additional \(400-\mathrm{g}\) load, the total force stretching the spring is \(7.84 \mathrm{~N}\). Then $$ y=\frac{F}{k}=\frac{7.84 \mathrm{~N}}{11.2 \mathrm{~N} / \mathrm{m}}=0.70 \mathrm{~m}=2 \times 35 \mathrm{~cm} $$ Provided it's Hookean, each \(400-\mathrm{g}\) load stretches the spring by the same amount, whether or not the spring is already loaded.

A \(20-\mathrm{kg}\) electric motor is mounted on four vertical springs, each having an elastic constant of \(30 \mathrm{~N} / \mathrm{cm}\). Find the period with which the motor vibrates vertically. As in Problem \(11.9\), we may replace the springs by an equivalent single spring. Its force constant will be \(4(3000 \mathrm{~N} / \mathrm{m})\) or \(12000 \mathrm{~N} / \mathrm{m}\). Then $$ T=2 \pi \sqrt{\frac{m}{k}}=2 \pi \sqrt{\frac{20 \mathrm{~kg}}{12000 \mathrm{~N} / \mathrm{m}}}=0.26 \mathrm{~s} $$

With a \(50-g\) mass at its end, a spring undergoes SHM with a frequency of \(0.70 \mathrm{~Hz}\). How much work is done in stretching the spring \(15 \mathrm{~cm}\) from its unstretched length? How much energy is then stored in the spring?

A \(50-\mathrm{g}\) 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 \((f)\) the acceleration when \(x=6.0 \mathrm{~cm}\). (a) \(f=\frac{1}{T}=\frac{1}{1.70 \mathrm{~s}}=0.588 \mathrm{~Hz}\) (b) Since \(T=2 \pi \sqrt{m / k}\) $$ k=\frac{4 \pi^{2} m}{T^{2}}=\frac{4 \pi^{2}(0.050 \mathrm{~kg})}{(1.70 \mathrm{~s})^{2}}=0.68 \mathrm{~N} / \mathrm{m} $$ (c) \(\left|v_{0}\right|=x_{0} \sqrt{\frac{k}{m}}=(0.12 \mathrm{~m}) \sqrt{\frac{0.68 \mathrm{~N} / \mathrm{m}}{0.050 \mathrm{~kg}}}=0.44 \mathrm{~m} / \mathrm{s}\) (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, $$ a_{0}=\frac{k}{m} x_{0}=\frac{0.68 \mathrm{~N} / \mathrm{m}}{0.050 \mathrm{~kg}}(0.12 \mathrm{~m})=1.6 \mathrm{~m} / \mathrm{s}^{2} $$ (e) From \(|v|=\sqrt{\left(x_{0}^{2}-x^{2}\right)(k / m)}\), $$ |v|=\sqrt{\frac{\left[(0.12 \mathrm{~m})^{2}-(0.06 \mathrm{~m})^{2}\right](0.68 \mathrm{~N} / \mathrm{m})}{(0.050 \mathrm{~kg})}}=0.38 \mathrm{~m} / \mathrm{s} $$ (f) \(a=-\frac{k}{m} x=-\frac{0.68 \mathrm{~N} / \mathrm{m}}{0.050 \mathrm{~kg}}(0.060 \mathrm{~m})=-0.82 \mathrm{~m} / \mathrm{s}^{2}\)

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?
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