Problem 13
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: Understand Simple Harmonic Motion
Simple harmonic motion can be described by the function \( x(t) = A \sin(2\pi ft) \) or \( x(t) = A \cos(2\pi ft) \), where \( A \) is the amplitude and \( f \) is the frequency. In this case, since the displacement is zero at \( t = 0 \), we will use the sine function.
2Step 2: Substitute Given Amplitude
The amplitude \( A \) is given as 6 inches. Substitute \( A = 6 \) into the equation to get: \( x(t) = 6 \sin(2\pi ft) \).
3Step 3: Substitute Given Frequency
The frequency \( f \) is given as \( \frac{5}{\pi} \) Hz. Substitute \( f = \frac{5}{\pi} \) into the equation: \( x(t) = 6 \sin\left(2\pi \cdot \frac{5}{\pi} t\right) \).
4Step 4: Simplify the Equation
Simplify the equation by multiplying inside the sine function: \( x(t) = 6 \sin(10t) \). This is the function that models the simple harmonic motion with the given properties.
Key Concepts
AmplitudeFrequencyTrigonometric Functions
Amplitude
In the context of simple harmonic motion, the amplitude represents the maximum displacement from the mean position. It shows how far the object moves from its equilibrium point. Amplitude is a key parameter to determine the extent of oscillation. It's usually represented by the letter \( A \) in the simple harmonic motion equations.
In our exercise, the amplitude is given as 6 inches. This means the maximum displacement, either in the positive or negative direction, is 6 inches from the equilibrium point. In simple terms, if you imagine a pendulum, the amplitude would be the furthest distance the pendulum swings from the center.
In our exercise, the amplitude is given as 6 inches. This means the maximum displacement, either in the positive or negative direction, is 6 inches from the equilibrium point. In simple terms, if you imagine a pendulum, the amplitude would be the furthest distance the pendulum swings from the center.
- Amplitude (\( A \)): 6 inches
- Represents maximum displacement
- Affects the height of the waveform
Frequency
Frequency in simple harmonic motion tells us how many cycles the motion completes in a specific time period, usually one second. It's an important measure because it indicates how fast an object is oscillating. In equations, frequency is represented by \( f \) and is measured in Hertz (\( \,Hz\)). One \( Hz \) corresponds to one cycle per second.
For the given exercise, the frequency is provided as \( \frac{5}{\pi}\) Hz. This value tells us that the motion completes \( \frac{5}{\pi} \) cycles every second. The formula for simple harmonic motion that includes frequency is \( x(t) = A \sin(2\pi ft) \), where \( 2\pi f \) determines the angular frequency. This directly affects how quickly the object moves through its cycle.
For the given exercise, the frequency is provided as \( \frac{5}{\pi}\) Hz. This value tells us that the motion completes \( \frac{5}{\pi} \) cycles every second. The formula for simple harmonic motion that includes frequency is \( x(t) = A \sin(2\pi ft) \), where \( 2\pi f \) determines the angular frequency. This directly affects how quickly the object moves through its cycle.
- Frequency (\( f \)): \( \frac{5}{\pi} \) Hz
- Indicates oscillation speed
- Higher frequency means more cycles per second
Trigonometric Functions
In simple harmonic motion, trigonometric functions like sine and cosine are used to describe the periodic nature of the movement. These functions are perfect for modeling oscillations because they inherently have a wave-like pattern, just like simple harmonic movements in real life.
Sine and cosine functions can model displacement over time, but the choice between them depends on the specific motion properties. In the given problem, we use the sine function \( x(t) = A \sin(2\pi ft) \) because the displacement is zero when \( t = 0 \). The sine function naturally starts from zero, which aligns perfectly with this initial condition.
Sine and cosine functions can model displacement over time, but the choice between them depends on the specific motion properties. In the given problem, we use the sine function \( x(t) = A \sin(2\pi ft) \) because the displacement is zero when \( t = 0 \). The sine function naturally starts from zero, which aligns perfectly with this initial condition.
- Sine function: Starts at zero, good for zero displacement at \( t = 0 \)
- Cosine function: Starts at a maximum, suitable for different starting conditions
- Trigonometric functions create repeatable wave patterns
Other exercises in this chapter
Problem 12
Find the period and graph the function. $$ y=\frac{1}{2} \tan x $$
View solution Problem 12
\(9-14\) . Find the missing coordinate of \(P\) , using the fact that \(P\) lies on the unit circle in the given quadrant. $$ \frac{\text { Coordinates }}{P\lef
View solution Problem 13
\(11-22\) . Use a calculator to find an approximate value of each expression correct to five decimal places, if it is defined. \(\cos ^{-1}\left(-\frac{3}{7}\ri
View solution Problem 13
Find the exact value of the trigonometric function at the given real number. (a) \(\sin \left(-\frac{\pi}{2}\right) \quad\) (b) \(\cos \left(-\frac{\pi}{2}\righ
View solution