A `20 cm` long rubber string fixed at both ends obeys Hook's law. Intially when it is stretched to make its total length of `24 cm`, the lowest frequency resonance is `v_(0)`. It is further stretched to make its total length of `26 cm`. The lowest frequency of resonance will now be :

Options

A. The same as `n_(0)`
B. Greater than `n_(0)`
C. Lower than `n_(0)`
D. None of these

Answer

B

Explanation

As we know
`n_(0) = (1)/(2L) sqrt((T)/(mu))`
Initially L = 25 cm , T = kx = k(4cm), `mu = (m)/(24)`
`n_(0) = (1)/(2 xx 24) sqrt((4k)/((m//24)) ) = 0.20 sqrt((k)/(m))`
When it is streched to the length 26 cm
L = 26 cm, T = k(6cm) , `mu = (m)/(26)`
`n_(0)^(') = (1)/(2 xx 26) sqrt((6k)/(m//26)) = 0.24 sqrt((k)/(m))`
`n_(0)^(') gt n_(0)`