Problem 78
Question
Sound Vibrations A tuning fork is struck, producing a pure tone as its tines vibrate. The vibrations are modeled by the function $$ v(t)=0.7 \sin (880 \pi t) $$ where \(v(t)\) is the displacement of the tines in millimeters at time \(t\) seconds. (a) Find the period of the vibration. (b) Find the frequency of the vibration, that is, the number of times the fork vibrates per second. (c) Graph the function \(v\)
Step-by-Step Solution
Verified Answer
(a) \( T = \frac{1}{440} \) seconds. (b) \( f = 440 \) Hz. (c) The graph is a sine wave with amplitude 0.7 and period \( \frac{1}{440} \) seconds.
1Step 1: Identify the General Form of the Sine Function
The given function is \( v(t) = 0.7 \sin(880\pi t) \). The general form of a sine function is \( y = A \sin(Bx) \), where \( A \) is the amplitude, \( B \) relates to the period of the function, and \( x \) is the variable.
2Step 2: Find the Period of the Vibration
The period \( T \) of a sine wave is given by the formula \( T = \frac{2\pi}{B} \). Here, \( B = 880\pi \). Substituting in the formula, we get \( T = \frac{2\pi}{880\pi} = \frac{1}{440} \). Hence, the period of the vibration is \( \frac{1}{440} \) seconds.
3Step 3: Find the Frequency of the Vibration
The frequency \( f \) is the number of cycles per second and is given by the inverse of the period, \( f = \frac{1}{T} \). From Step 2, we know the period \( T = \frac{1}{440} \), so the frequency \( f = 440 \) Hz. Therefore, the tuning fork vibrates 440 times per second.
4Step 4: Graph the Function
To graph \( v(t) = 0.7 \sin(880\pi t) \), plot the sine wave with an amplitude of 0.7, a period of \( \frac{1}{440} \), and no vertical shift. The x-axis represents time \( t \) in seconds, and y-axis represents the displacement \( v(t) \) in millimeters. The wave will complete one cycle from its peak, through zero, to its trough, and back to zero within \( \frac{1}{440} \) seconds.
Key Concepts
Period of a FunctionFrequency of VibrationsSine Wave Graphing
Period of a Function
In mathematics, the period of a function refers to the length of the interval after which the function repeats itself. For trigonometric functions like sine functions, this concept is vital as it describes the repeating nature of these waves.
The function given in the exercise is modeled as \( v(t) = 0.7 \sin(880\pi t) \). This is a standard sine wave where you can identify the period using the coefficient \( B \) from the general sine function form \( y = A \sin(Bx) \).
The function given in the exercise is modeled as \( v(t) = 0.7 \sin(880\pi t) \). This is a standard sine wave where you can identify the period using the coefficient \( B \) from the general sine function form \( y = A \sin(Bx) \).
- The formula to find the period \( T \) is \( T = \frac{2\pi}{B} \), where \( B \) in this case is \( 880\pi \).
- Upon substituting, we calculate \( T = \frac{2\pi}{880\pi} = \frac{1}{440} \).
Frequency of Vibrations
Frequency is a measure of how often a repeating event occurs per unit of time. In the context of sound waves, it measures how many vibrations occur in one second, defining the pitch of the sound produced.
To find the frequency of the vibration from the given function \( v(t) = 0.7 \sin(880\pi t) \), we use the relationship between period and frequency.
To find the frequency of the vibration from the given function \( v(t) = 0.7 \sin(880\pi t) \), we use the relationship between period and frequency.
- The frequency \( f \) is the reciprocal of the period, expressed as \( f = \frac{1}{T} \).
- Given that the period \( T \) is \( \frac{1}{440} \), the frequency \( f = 440 \) Hz.
Sine Wave Graphing
Graphing a sine wave like the one described by the function \( v(t) = 0.7 \sin(880\pi t) \) reveals the oscillating nature of the wave.
To graph this function, follow these steps:
To graph this function, follow these steps:
- Recognize the function has an amplitude of 0.7, meaning the wave reaches 0.7 millimeters above and below the x-axis.
- Identify the period as \( \frac{1}{440} \) seconds, showing how quickly the wave repeats within a second.
- The starting point is zero, and the wave moves to its maximum, back through zero to the minimum, and returns to zero, completing one full cycle.
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