Question

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

Answer 1

`Q=50.3`

Step-by-step explanation:

From the question we are told that:

Frequency
`F=330Hz`

Sound intensity drop
`I_d=2`

Time
`T=4s`

Therefore

Sound intensity Ratio

`(I)/(I_x)=(1)/(2)`

Generally the equation for Sound intensity is mathematically given by

`(I)/(I_x)=e^(-4\ \=t)`

`(1)/(2)=e^(-4\ \=t)`

`\=t =5.8s`

Generally the equation for Quality Factor is mathematically given by

`Q=2 \pi(E)/(\triangle E)`

`Q=2 \pi(E)/((E)/(2*4))`

`Q=50.3`