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An object is undergoing SHM with period 1.200 s and amplitude 0.600 m. At \(t =\) 0 the object is at \(x =\) 0 and is moving in the negative \(x\)-direction. How far is the object from the equilibrium position when \(t =\) 0.480 s?

Short Answer

Expert verified
The object is approximately 0.571 m from the equilibrium position at t = 0.480 s.

Step by step solution

01

Formula for SHM Displacement

The equation for the displacement of an object in simple harmonic motion (SHM) is given by:\[ x(t) = A \cos(\omega t + \phi) \]where:- \( A \) is the amplitude.- \( \omega \) is the angular frequency.- \( \phi \) is the phase constant.Given that \( A = 0.600 \) m and the period \( T = 1.200 \) s, we need to find \( \omega \).
02

Calculate Angular Frequency

The angular frequency \( \omega \) can be calculated using the formula:\[ \omega = \frac{2\pi}{T} \]Substitute \( T = 1.200 \) s:\[ \omega = \frac{2\pi}{1.200} = \frac{5\pi}{3} \text{ rad/s} \]
03

Determine Phase Constant φ

At \( t = 0 \), the object is at \( x = 0 \) and moving in the negative x-direction, which suggests a sine function is more suitable for this condition. Therefore, we use:\[ x(t) = A \sin(\omega t) \]However, initially using cosine, for consistency, we have to use:\[ x(0) = A \cos(\phi) = 0 \]which implies \( \phi = \frac{\pi}{2} \) since \( \cos(\frac{\pi}{2}) = 0 \). Adjusting the equation indicates the object is indeed moving in the negative direction.
04

Apply SHM Equation at t = 0.480 s

Substitute \( t = 0.480 \) s and the values \( A = 0.600 \), \( \omega = \frac{5\pi}{3} \), \( \phi = \frac{\pi}{2} \) into the displacement equation:\[ x(0.480) = 0.600 \cos\left(\frac{5\pi}{3} \times 0.480 + \frac{\pi}{2}\right) \]
05

Simplify and Calculate

Calculate the inside of the cosine:\[ \frac{5\pi}{3} \times 0.480 = 0.800\pi \]Then add the phase constant:\[ 0.800\pi + \frac{\pi}{2} = 0.800\pi + 0.500\pi = 1.300\pi \]Now evaluate the cosine:\[ x(0.480) = 0.600 \cos(1.300\pi) = 0.600 \cos(\pi + 0.300\pi) = 0.600 (-\cos(0.300\pi)) \]Where \( \cos(0.300\pi) \approx 0.9511 \), so,\[ x(0.480) = -0.600 \times 0.9511 \approx -0.5707 \text{ m} \]
06

Find Distance from Equilibrium

The distance from the equilibrium position is the absolute value of the displacement:\[ |x(0.480)| = |-0.5707| \approx 0.5707 \text{ m} \]

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Key Concepts

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

Amplitude
In simple harmonic motion (SHM), the amplitude represents the maximum displacement of the object from its equilibrium position. In essence, it illustrates how far the object moves on either side of this central point.
In our exercise, the amplitude is given as 0.600 m. This means the object can move up to 0.600 meters away from the equilibrium before changing directions.
  • The greater the amplitude, the larger the displacement of the object from equilibrium.
  • Amplitude is crucial as it influences the energy of the system. A higher amplitude means more energy.
When solving problems involving SHM, always remember that while amplitude affects the total energy, it does not influence the period or frequency of the motion.
Angular Frequency
Angular frequency (\(\omega\)) in SHM is a measure of how quickly the object moves through its cycle. It is related to the oscillation period and is expressed in radians per second.
To find angular frequency, use the equation:\[\omega = \frac{2\pi}{T}\]where \(T\) is the period of oscillation. In this exercise, with \(T = 1.200\) s, the angular frequency calculates to:\[\omega = \frac{2\pi}{1.200} = \frac{5\pi}{3}\, \text{rad/s}\]Ul bullet points can be really helpful here!
  • Angular frequency helps determine the speed of oscillations.
  • It's independent of the amplitude but crucial for understanding how cycles progress over time.
Thus, learn always to calculate the angular frequency to comprehend better how fast the system oscillates.
Phase Constant
The phase constant (\(\phi\)) in the context of SHM determines the starting position of the oscillating object relative to its equilibrium. It effectively shifts the wave horizontally on a graph.
Our exercise provides insights into why determining the phase constant is essential. Since the object starts at \(x = 0\) and moves negatively, we use the cosine form,\[A \cos(\phi) = 0\]resulting in \(\phi = \frac{\pi}{2}\).Switching functions or solving for \(\phi\) depends on initial conditions:
  • The phase constant is often chosen to match initial displacement and velocity conditions.
  • A correct phase constant aligns theoretical models to actual movements.
Understanding phase constants allow for more accurate predictions of position over time.
Cosine Function
In SHM, the cosine function is often used to model the displacement of an object over time. The function:\[x(t) = A \cos(\omega t + \phi)\]captures how position changes with time using amplitude, angular frequency, and phase constant.
In this exercise, translating theoretical equations to practical values involved:
  • Substituting given quantities such as amplitude, angular frequency, and phase constant.
  • Evaluating the equation at specific times to find its current position.
Ultimately, cosine functions are pivotal for calculating how far an object is from equilibrium at any given moment, making it central to analyzing SHM.

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Most popular questions from this chapter

A spring of negligible mass and force constant \(k =\) 400 N/m is hung vertically, and a 0.200-kg pan is suspended from its lower end. A butcher drops a 2.2-kg steak onto the pan from a height of 0.40 m. The steak makes a totally inelastic collision with the pan and sets the system into vertical SHM. What are (a) the speed of the pan and steak immediately after the collision; (b) the amplitude of the subsequent motion; (c) the period of that motion?

A machine part is undergoing SHM with a frequency of 4.00 Hz and amplitude 1.80 cm. How long does it take the part to go from \(\chi =\) 0 to \(\chi = -1.80\) cm?

(a) \(\textbf{Music}\). When a person sings, his or her vocal cords vibrate in a repetitive pattern that has the same frequency as the note that is sung. If someone sings the note B flat, which has a frequency of 466 Hz, how much time does it take the person's vocal cords to vibrate through one complete cycle, and what is the angular frequency of the cords? (b) \(\textbf{Hearing}\). When sound waves strike the eardrum, this membrane vibrates with the same frequency as the sound. The highest pitch that young humans can hear has a period of 50.0 \(\mu\)s. What are the frequency and angular frequency of the vibrating eardrum for this sound? (c) \(\textbf{Vision}\). When light having vibrations with angular frequency ranging from 2.7 \(\times\) 10\(^{15}\) rad/s to 4.7 \(\times\) 10\(^{15}\) rad/s strikes the retina of the eye, it stimulates the receptor cells there and is perceived as visible light. What are the limits of the period and frequency of this light? (d) \(\textbf{Ultrasound}\). High frequency sound waves (ultrasound) are used to probe the interior of the body, much as x rays do. To detect small objects such as tumors, a frequency of around 5.0 MHz is used. What are the period and angular frequency of the molecular vibrations caused by this pulse of sound?

A 1.80-kg monkey wrench is pivoted 0.250 m from its center of mass and allowed to swing as a physical pendulum. The period for small-angle oscillations is 0.940 s. (a) What is the moment of inertia of the wrench about an axis through the pivot? (b) If the wrench is initially displaced 0.400 rad from its equilibrium position, what is the angular speed of the wrench as it passes through the equilibrium position?

The balance wheel of a watch vibrates with an angular amplitude \(\Theta\), angular frequency \(\omega\), and phase angle \(\phi =\) 0. (a) Find expressions for the angular velocity \(d\theta/dt\) and angular acceleration \(d^2\theta/dt^2\) as functions of time. (b) Find the balance wheel's angular velocity and angular acceleration when its angular displacement is \(\Theta\), and when its angular displacement is \(\Theta\)/2 and \(\theta\) is decreasing. (\(Hint\): Sketch a graph of \(\theta\) versus \(t\).)

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