SOLUTION: A 6-string guitar is being strummed by a guitarist once. When he strummed the strings, he hit the 6th string (topmost string) first and released it before hitting the 5th string co

Question 1175988: A 6-string guitar is being strummed by a guitarist once. When he strummed the strings, he hit the 6th string (topmost string) first and released it before hitting the 5th string consequently. The 6th string went back to its initial position after 50 milliseconds from its initial position. When the 6th string reaches its maximum point, the 5th string went back to its initial position for the first time. Assume an equal tension applied to both strings.
A. Find the equation of the 6th string and graph.
B. Find the equation of the 5th string and graph.
1. Assume that the 5th string is released when the 6th string went back to its initial position for the first time.
2. Assume that the oscillation of the 5th string started when the 6th string reached its maximum point.

Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
You've asked this question multiple times. Let's reiterate the solution with a focus on clarity.
**Assumptions:**
* Simple harmonic motion (sinusoidal wave).
* Initial position is equilibrium (y = 0).
* Equal amplitude (A) for both strings.
* Frequency is determined by the time to return to the initial position.
**A. Equation of the 6th String and Graph**
1. **Period (T):**
* T = 50 milliseconds = 0.05 seconds.
2. **Angular Frequency (ω):**
* ω = 2π / T = 2π / 0.05 = 40π radians/second.
3. **Amplitude (A):**
* Let A be the amplitude (remains undefined).
4. **Phase Shift (φ):**
* Starts at y = 0 at t = 0, so φ = 0.
5. **Equation:**
* y6(t) = A * sin(ωt + φ) = A * sin(40πt).
6. **Graph:**
* A sine wave with a period of 0.05 seconds, starting at (0, 0).
**B. Equation of the 5th String and Graph**
**1. Released when the 6th string returns to its initial position.**
1. **Start Time:**
* t = 0.05 seconds.
2. **Period and Angular Frequency:**
* T = 0.05 seconds, ω = 40π radians/second.
3. **Phase Shift:**
* y5(t) = A * sin(40π(t - 0.05)) = A * sin(40πt - 2π) = A * sin(40πt).
4. **Equation:**
* y5(t) = A * sin(40πt).
5. **Graph:**
* Identical to the 6th string's graph.
**2. Released when the 6th string reaches its maximum point.**
1. **Time of Maximum Point:**
* t = T/4 = 0.05/4 = 0.0125 seconds.
2. **Period and Angular Frequency:**
* T = 0.05 seconds, ω = 40π radians/second.
3. **Phase Shift:**
* y5(t) = A * sin(40π(t - 0.0125)) = A * sin(40πt - π/2) = -A * cos(40πt).
4. **Equation:**
* y5(t) = -A * cos(40πt).
5. **Graph:**
* A negative cosine wave with a period of 0.05 seconds, starting at (0, -A).
**Summary:**
* **6th String:** y6(t) = A * sin(40πt)
* **5th String (1):** y5(t) = A * sin(40πt)
* **5th String (2):** y5(t) = -A * cos(40πt)
**Key Points:**
* The amplitude (A) is a variable.
* The period (T) is 0.05 seconds.
* The phase shift determines the 5th string's starting point.

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