Solution:** To find the time \( t \) when the frequency \( f(t) = 20 \), we set up the equation:
["# How to Find the Time ( t ) When Frequency ( f(t) = 20 ): A Clear Solution", "Understanding when a periodic function reaches a specific frequency value is essential in fields like engineering, physics, and signal processing. In this article, we’ll explore how to determine the time ( t ) at which the frequency ( f(t) ) equals 20 by setting up and solving the appropriate equation.", "## Understanding Frequency in Time-Domain Functions", "Frequency ( f(t) ) typically represents how often a signal oscillates within a function ( f(t) ) at time ( t ). For sinusoidal or harmonic functions, frequency is often constant. However, in dynamic systems or time-varying signals, ( f(t) ) can change, meaning we must solve an equation involving ( t ) to find when ( f(t) = 20 ).", "## Setting Up the Equation ( f(t) = 20 )", "To find the time ( t ) when ( f(t) = 20 ), begin by writing the explicit form of ( f(t) ). Suppose the function is modeled as:", "[\nf(t) = A \sin(Bt + C) + D\n]", "Where:\n- ( A ) is the amplitude,\n- ( B ) relates to angular frequency (( B = 2\pi f_0 )),\n- ( C ) is the phase shift,\n- ( D ) is the vertical shift.", "Set ( f(t) = 20 ):", "[\nA \sin(Bt + C) + D = 20\n]", "Subtract ( D ) from both sides:", "[\nA \sin(Bt + C) = 20 - D\n]", "Divide by ( A ):", "[\n\sin(Bt + C) = \frac{20 - D}{A}\n]", "For a solution to exist, the right-hand side must lie within ([-1, 1]). Assuming this condition is met, solve the equation:", "[\nBt + C = \arcsin\left( \frac{20 - D}{A} \right) + 2\pi n \quad \ ext{or} \quad \pi - \arcsin\left( \frac{20 - D}{A} \right) + 2\pi n\n]", "where ( n ) is any integer. Solve for ( t ):", "[\nt = \frac{1}{B} \left[ \arcsin\left( \frac{20 - D}{A} \right) + C + 2\pi n \right]\n\quad \ ext{or} \quad\nt = \frac{1}{B} \left[ \pi - \arcsin\left( \frac{20 - D}{A} \right) + C + 2\pi n \right]\n]", "## Practical Steps to Solve", "1. Identify the function form: Know whether ( f(t) ) is sinusoidal, exponential, or defined by differential equations.\n2. Isolate ( f(t) = 20 ): Rearrange the equation to express its dependency on ( t ).\n3. Use inverse trigonometric functions: Since frequency is often tied to sine or cosine dynamics, apply arcsine or arccosine as needed.\n4. Incorporate periodicity: Include integer ( n ) to find all times ( t ) satisfying the condition.\n5. Validate solutions: Ensure ( \left| \frac{20 - D}{A} \right| \leq 1 ) for real solutions.", "## Example", "Suppose ( f(t) = 5 \sin(2\pi t) + 10 ). Find ( t ) when ( f(t) = 20 ):", "[\n5 \sin(2\pi t) + 10 = 20 \implies \sin(2\pi t) = 2\n]", "Since ( \sin(\ heta) \leq 1 ), no real ( t ) satisfies this. So, adjust parameters—say ( f(t) = 15 \sin(2\pi t) + 5 ). Then:", "[\n15 \sin(2\pi t) + 5 = 20 \implies \sin(2\pi t) = 1\n]", "[\n2\pi t = \frac{\pi}{2} + 2\pi n \implies t = \frac{1}{4} + n\n]", "Thus, ( t = 0.25, 1.25, 2.25, \dots ) are solutions.", "## Conclusion", "Finding the time ( t ) when ( f(t) = 20 ) involves forming and solving an equation that captures how frequency depends on time—often through inverse sine functions. By isolating ( t ) and accounting for periodicity, you can determine all moments when the signal reaches the target frequency. This method is foundational for analyzing oscillating systems and optimizing dynamic signals.", "### Key Takeaways:\n- Frequency equations often use trigonometric relationships.\n- Solving ( f(t) = k ) leads to inverse trigonometric solutions.\n- Periodicity extends solutions across time via modular arithmetic.\n- Always check domain constraints for realistic solutions.", "Use this framework whenever analyzing frequency behavior in mathematical or applied contexts—accuracy begins with precise equation setup."]
