The length and diameter of a metal wire is doubled the fundamental frequency of vibration will change from n to (tension being kept constant and material of both the wires is same)

Options

A. `n/4`
B. `n/8`
C. `n/12`
D. `n/16`

Answer

a

Explanation

Frequency of vibration in the wire,
`v=1/(2l)sqrt(T/mu)` … (i)
where ,` l=` length of the wire
T = tension in the wire
`mu=` mass per unit length
`v'=1/(2l')sqrt(T/mu)` …(ii)
From Eqs. (i) and (ii),
`v/(v') = (l')/lxx sqrt((mu')/mu) = (2l)/1 xx sqrt((ppir'^(2))/(ppir^(2)))`
`= (2l)/lxx sqrt(((r')/r)^(2))=2xx(r')/r=2xx2=4`
`rArr v= n= 4v' rArr v'= (n)/4`