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Magazine Chapter 14: Problem 65
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 蠁
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.
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.
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!
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.
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:
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.
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:
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.
