Problem 11
Question
Find a function that models the simple harmonic motion having the given properties. Assume that the displacement is zero at time \(t=0\) . amplitude 6 in., frequency 5\(/ \pi \mathrm{Hz}\)
Step-by-Step Solution
Verified Answer
The function is \( x(t) = 6 \sin(10t) \).
1Step 1: Understanding Simple Harmonic Motion
In simple harmonic motion, displacement as a function of time is typically modeled using the equation \( x(t) = A \cos(\omega t + \phi) \) or \( x(t) = A \sin(\omega t + \phi) \). Here, \( A \) is the amplitude, \( \omega \) is the angular frequency in radians per second, \( \phi \) is the phase constant, and \( t \) is the time.
2Step 2: Determine Angular Frequency
The frequency \( f \) is given as \( \frac{5}{\pi} \) Hz. Angular frequency \( \omega \) is calculated as \( \omega = 2\pi f \). Substituting in the given frequency, \( \omega = 2\pi \times \frac{5}{\pi} = 10 \) radians per second.
3Step 3: Set Phase Constant for Initial Condition
At \( t = 0 \), displacement \( x(t) \) is zero. Therefore, for the sine function, \( x(0) = A \sin(\phi) = 0 \). The simplest solution for this is \( \phi = 0 \). Thus, the equation simplifies to \( x(t) = 6 \sin(10t) \) since the amplitude \( A = 6 \) inches.
Key Concepts
Angular FrequencyAmplitudePhase Constant
Angular Frequency
Angular frequency, often represented by the symbol \( \omega \), is a key concept in simple harmonic motion. It describes how quickly the oscillating object moves through its cycle and is measured in radians per second. To find the angular frequency from a given frequency \( f \), you use the formula \( \omega = 2\pi f \). This formula relates linear frequency, measured in hertz (Hz), to angular frequency.
For example, if the frequency of oscillation is given as \( \frac{5}{\pi} \) Hz, calculate the angular frequency by substituting into the formula:
For example, if the frequency of oscillation is given as \( \frac{5}{\pi} \) Hz, calculate the angular frequency by substituting into the formula:
- \( \omega = 2\pi \times \frac{5}{\pi} \)
- After simplification, \( \omega = 10 \) radians per second.
Amplitude
Amplitude, denoted by \( A \), is the maximum displacement of an object from its rest position during oscillation. In simple harmonic motion, amplitude represents the peak value of displacement or how "far" and to what "extent" the motion occurs from the central point.
In our exercise, the amplitude is provided as 6 inches. This means:
In our exercise, the amplitude is provided as 6 inches. This means:
- The object will move 6 inches away from its equilibrium or central position, at maximum, on either side throughout its motion.
- Amplitude determines the "height" of the oscillation in the model function of simple harmonic motion.
Phase Constant
The phase constant \( \phi \) is a parameter in the equations describing simple harmonic motion, namely \( x(t) = A \sin(\omega t + \phi) \) or \( x(t) = A \cos(\omega t + \phi) \). This constant shifts the function along the time axis, indicating where the motion starts at \( t = 0 \).
In our problem, the displacement is zero at \( t = 0 \), which indicates that using the sine function is appropriate since \( \sin(0) = 0 \). Thus, the phase constant \( \phi \) should be 0 to satisfy the given conditions:
In our problem, the displacement is zero at \( t = 0 \), which indicates that using the sine function is appropriate since \( \sin(0) = 0 \). Thus, the phase constant \( \phi \) should be 0 to satisfy the given conditions:
- \( x(0) = A \sin(0 + \phi) \)
- For the displacement to be zero, \( \phi \) must also be zero in this case.
Other exercises in this chapter
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