The strings of a violin are tuned to the tones `G , D , A and E` which are separated by a fifth from one another . That is `f(D) = 1.5 (G) , f(A) = 1.5 f(D) = 400 Hz and f(E ) = 1.5 f(A)`. The distance between the two fixed points , the bridge at the scroll and over the body of the instrument is `0.25 m`. The tension on the string `E is 90 N`. The mass per unit length of string `E` is nearly

Options

A. ` 1 g//m`
B. ` 2 g//m`
C. ` 3 g// m`
D. `4 g//m`

Answer

A

Explanation

`f ( E ) = 1.5 xx 400 = 600 Hz = (1)/( 2 l) sqrt((T)/(mu))`
` = (1)/( 2 xx 0.25) sqrt((90)/(mu))`
`mu = (90)/((0.5)^(2) xx (600)^(2)) = 1 g//m`