A string is stretched betweeb fixed points separated by `75.0 cm`. It observed to have resonant frequencies of `420 Hz` and `315 Hz`. There are no other resonant frequencies between these two. The lowest resonant frequency for this strings is

Answer

(a) `105 Hz`; (b) `157.5 m//s`

Explanation

`F_(1) = 420 = (nv)/(2L)`
`F_(2) = 3.15 = (mv)/(2L)`
`(F_(1))/(F_(2)) = (420)/(315) = (n)/(m)` , (here, `n` and `m` are integers)
So `n = 4 , m = 3`
No any resonant frequency so
`420 = (4v)/(2L)`
`(v)/(2L) = ((420)/(2L)) = 105 H_(2)` (Lowest resonant frequency.)
`V = (105)(2L) = 105 xx 2 xx 0.75 = 105 xx (3)/(2) = 157.5 m//sec`.