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When the E string of a guitar (frequency 330 Hz) is plucked, the sound intensity decreases by a factor of 2 after 4 s. Determine

Sagot :

okpalawalter8

Answer:

[tex]Q=50.3[/tex]

Explanation:

From the question we are told that:

Frequency [tex]F=330Hz[/tex]

Sound intensity drop [tex]I_d=2[/tex]

Time [tex]T=4s[/tex]

Therefore

Sound intensity Ratio

 [tex]\frac{I}{I_x}=\frac{1}{2}[/tex]

Generally the equation for Sound intensity is mathematically given by

 [tex]\frac{I}{I_x}=e^{-4\ \=t}[/tex]

 [tex]\frac{1}{2}=e^{-4\ \=t}[/tex]

 [tex]\=t =5.8s[/tex]

Generally the equation for Quality Factor is mathematically given by

 [tex]Q=2 \pi\frac{E}{\triangle E}[/tex]

 [tex]Q=2 \pi\frac{E}{\frac{E}{2*4}}[/tex]

 [tex]Q=50.3[/tex]

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