5 Type the correct answer in each box. Use numerals instead of words. A tuning fork vibrates with a frequency of 440 hertz (cycles/second). When the tuning fork is struck, it produces a change in the room. Function \( p \) represents this situation, where \( p(t) \) is the change in pressure, in pascals, relative to the normal air pressure in th time, \( t \), in seconds, after the tuning fork is struck. What are the domain \( (880 \pi t) \) The domain the range within the context of this situation? The range of the function is \( \square \leq p(t) \leq \)

To analyze the situation with the tuning fork vibrating at a frequency of 440 Hz, we can derive the function \( p(t) \) that represents the change in pressure over time. 1. **Frequency and Angular Frequency**: The frequency \( f \) is given as 440 Hz. The angular frequency \( \omega \) can be calculated using the formula: \[ \omega = 2\pi f = 2\pi \times 440 = 880\pi \text{ radians/second} \] 2. **Function Representation**: The change in pressure \( p(t) \) can be modeled as a sinusoidal function, typically a sine or cosine function. A common representation is: \[ p(t) = A \sin(880\pi t) \] where \( A \) is the amplitude of the pressure change. 3. **Domain**: The domain of the function \( p(t) \) in this context is typically all non-negative time values since we are measuring time after the tuning fork is struck. Therefore, the domain is: \[ t \geq 0 \] 4. **Range**: The range of the function depends on the amplitude \( A \). The sine function oscillates between -1 and 1, so the range of \( p(t) \) will be: \[ -A \leq p(t) \leq A \] Thus, if we denote the amplitude as \( A \), the range can be expressed as: \[ -A \leq p(t) \leq A \] In summary, the answers are: - **Domain**: \( t \geq 0 \) - **Range**: \( -A \leq p(t) \leq A \) (where \( A \) is the amplitude of the pressure change).