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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 1.50-kg, horizontal, uniform tray is attached to a vertical ideal spring of force constant 185 N/m and a 275-g metal ball is in the tray. The spring is below the tray, so it can oscillate up and down. The tray is then pushed down to point \(A\), which is 15.0 cm below the equilibrium point, and released from rest. (a) How high above point \(A\) will the tray be when the metal ball leaves the tray? (\(Hint\): This does \(not\) occur when the ball and tray reach their maximum speeds.) (b) How much time elapses between releasing the system at point \(A\) and the ball leaving the tray? (c) How fast is the ball moving just as it leaves the tray?

A 2.00-kg bucket containing 10.0 kg of water is hanging from a vertical ideal spring of force constant 450 N/m and oscillating up and down with an amplitude of 3.00 cm. Suddenly the bucket springs a leak in the bottom such that water drops out at a steady rate of 2.00 g/s. When the bucket is half full, find (a) the period of oscillation and (b) the rate at which the period is changing with respect to time. Is the period getting longer or shorter? (c) What is the shortest period this system can have?

An apple weighs 1.00 N. When you hang it from the end of a long spring of force constant 1.50 N/m and negligible mass, it bounces up and down in SHM. If you stop the bouncing and let the apple swing from side to side through a small angle, the frequency of this simple pendulum is half the bounce frequency. (Because the angle is small, the back-and-forth swings do not cause any appreciable change in the length of the spring.) What is the unstretched length of the spring (with the apple removed)?

A 0.0200-kg bolt moves with SHM that has an amplitude of 0.240 m and a period of 1.500 s. The displacement of the bolt is \(+\)0.240 m when \(t =\) 0. Compute (a) the displacement of the bolt when \(t =\) 0.500 s; (b) the magnitude and direction of the force acting on the bolt when \(t =\) 0.500 s; (c) the minimum time required for the bolt to move from its initial position to the point where \(x = -\)0.180 m; (d) the speed of the bolt when \(x = -\)0.180 m.

A 10.0-kg mass is traveling to the right with a speed of 2.00 m/s on a smooth horizontal surface when it collides with and sticks to a second 10.0-kg mass that is initially at rest but is attached to a light spring with force constant 170.0 N/m. (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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