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Magazine Chapter 14: Problem 5
A machine part is undergoing SHM with a frequency of 5.00 \(\mathrm{Hz}\) and amplitude 1.80 \(\mathrm{cm} .\) How long does it take the part to go from \(x=0\) to \(x=-1.80 \mathrm{cm} ?\)
Short Answer
Expert verified
The part takes 0.15 seconds to go from \(x = 0\) to \(x = -1.80 \mathrm{cm}\).
Step by step solution
01
Understand Simple Harmonic Motion (SHM)
In SHM, an oscillating object moves back and forth about an equilibrium position. The motion can be described by \[ x(t) = A \cos(\omega t + \phi) \] where \(A\) is the amplitude, \(\omega\) is the angular frequency, \(t\) is time, and \(\phi\) is the phase constant.
02
Establish Given Values
We know the amplitude \(A = 1.80\, \mathrm{cm}\) and frequency \(f = 5.00\, \mathrm{Hz}\). From the frequency, we calculate the angular frequency with \[ \omega = 2\pi f \].
03
Calculate Angular Frequency
Calculate the angular frequency: \[ \omega = 2 \pi \times 5.00\, \mathrm{Hz} = 10\pi\, \mathrm{rad/s} \].
04
Determine Initial Conditions
Assume the motion starts at \(x = 0\) when \(t = 0\), which implies the phase constant \( \phi = -\frac{\pi}{2} \) (since cosine is zero at -蟺/2). Therefore, the position equation becomes \[ x(t) = 1.80 \cos\left(10\pi t - \frac{\pi}{2}\right) \].
05
Set the Position Equation
We want the object to go from \(x = 0\) to \(x = -1.80\, \mathrm{cm}\). So set \[ -1.80 = 1.80 \cos\left(10\pi t - \frac{\pi}{2}\right) \].
06
Solve the Equation for Time
Divide both sides by 1.80, then take the inverse cosine: \[ \cos\left(10\pi t - \frac{\pi}{2}\right) = -1 \] means \[ 10\pi t - \frac{\pi}{2} = \pi \]. Solve for \(t\): \[ 10\pi t = \pi + \frac{\pi}{2} = \frac{3\pi}{2} \] thus \[ t = \frac{3}{20}\, \mathrm{s} = 0.15\, \mathrm{s} \].
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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, amplitude plays a crucial role in understanding the extent of motion. Amplitude, denoted by the letter \( A \), refers to the maximum displacement from the equilibrium position. It is the farthest point the object moves from its central position before it reverses direction.
- This measure is always a positive value, regardless of the direction of displacement.
- In our given problem, the amplitude is \( 1.80 \) cm, which means the furthest distance the machine part moves from its central position is 1.80 cm.
- Amplitude does not affect the period or frequency of the oscillation, but it does affect the energy of the oscillating system.
Angular Frequency
Angular frequency is an essential concept in the study of oscillations, particularly simple harmonic motion. Represented by \( \omega \), angular frequency indicates how fast an object is oscillating through its cycle.
- It is calculated as \( \omega = 2\pi f \), where \( f \) is the frequency of oscillation.
- In the provided exercise, the compute angular frequency is \( 10\pi \) rad/s, derived from the given frequency of 5.00 Hz.
- Angular frequency is crucial in determining the speed of oscillations, and it also appears in the equations of motion, as it influences the time period of oscillations.
Phase Constant
The phase constant \( \phi \) is a parameter in the simple harmonic motion equation that adjusts the initial start position of the oscillating object. It defines where in its cycle the oscillation begins at \( t = 0 \).
- For instance, a phase constant of zero means the object starts at maximum displacement, while other values indicate different starting positions.
- In our solution, the motion starts at \( x = 0 \), implying that the phase constant is \( -\frac{\pi}{2} \), resulting in the cosine term being zero, aligning with the initial condition.
- The phase constant is vital for aligning the mathematical model with actual initial conditions of motion.
Frequency
Frequency, denoted as \( f \), indicates how many oscillations or cycles occur in a given period, typically one second. Measured in hertz (Hz), it is a fundamental property of any repeating motion, including simple harmonic motion.
- Higher frequency means more oscillations per second, while lower frequency means fewer.
- The exercise indicates a frequency of \( 5.00 \) Hz, meaning the machine part completes 5 oscillations every second.
- Frequency and period are reciprocally related: \( f = \frac{1}{T} \), where \( T \) is the period, the duration of one complete oscillation cycle.
