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

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Question 1176289: 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 normal 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.
Found 2 solutions by CPhill, ikleyn:
Answer by CPhill(1959) (Show Source):
You can put this solution on YOUR website!
Let's break down this problem step-by-step.
Assumptions:
The strings oscillate with a simple harmonic motion (sinusoidal wave).
The initial position of the strings is considered the equilibrium position (y = 0).
The amplitude of the oscillations is the same for both strings due to equal tension.
The frequency of oscillation is determined by the time it takes for a string to return to its initial position.
A. Find the Equation of the 6th String and Graph
Period (T):
The 6th string returns to its initial position after 50 milliseconds (0.05 seconds). This is the period of oscillation.
T = 0.05 seconds
Angular Frequency (ω):
ω = 2π / T
ω = 2π / 0.05 = 40π radians/second
Amplitude (A):
We don't have the amplitude, so let's call it A. We'll assume A is positive.
Phase Shift (φ):
The 6th string is hit first, so we'll assume it starts at its initial position (y = 0) at t = 0. This means there's no phase shift (φ = 0).
Equation of the 6th String (y6(t)):
y6(t) = A * sin(ωt + φ)
y6(t) = A * sin(40πt)
Graph:
The graph will be a sine wave with a period of 0.05 seconds.
The amplitude will be A.
The wave will start at y = 0 at t = 0.
B. Find the Equation of the 5th String and Graph
1. Assume the 5th string is released when the 6th string went back to its initial position for the first time.
Start Time:
The 5th string starts oscillating at t = 0.05 seconds.
Period and Angular Frequency:
We assume the 5th string has the same period and angular frequency as the 6th string:
T = 0.05 seconds
ω = 40π radians/second
Phase Shift:
To find the phase shift, we need to shift the sine wave by 0.05 seconds.
y5(t) = A * sin(40π(t - 0.05))
y5(t) = A * sin(40πt - 2π)
Since sin(x - 2π) = sin(x), we have:
y5(t) = A * sin(40πt)
Equation of the 5th String (y5(t)):
y5(t) = A * sin(40πt)
Graph:
The graph will be identical to the 6th string's graph.
2. Assume the oscillation of the 5th string started when the 6th string reached its maximum point.
Time of Maximum Point:
The 6th string reaches its maximum at t = T/4 = 0.05/4 = 0.0125 seconds.
Period and Angular Frequency:
Same as before:
T = 0.05 seconds
ω = 40π radians/second
Phase Shift:
We need to shift the sine wave by 0.0125 seconds.
y5(t) = A * sin(40π(t - 0.0125))
y5(t) = A * sin(40πt - π/2)
Since sin(x - π/2) = -cos(x), we have:
y5(t) = -A * cos(40πt)
Equation of the 5th String (y5(t)):
y5(t) = -A * cos(40πt)
Graph:
The graph will be a negative cosine wave with a period of 0.05 seconds.
The amplitude will be A.
The wave will start at y = -A at t = 0.
Summary:
6th String: y6(t) = A * sin(40πt)
5th String (1): y5(t) = A * sin(40πt)
5th String (2): y5(t) = -A * cos(40πt)
Note: The amplitude (A) remains undefined in this problem, but it would be determined by the initial energy imparted to the strings.

Answer by ikleyn(52756) (Show Source):
You can put this solution on YOUR website!