A steel wire of length 60 cm and area of cross-section `10^(-6) m^(2)` is joined with an aluminium wire of length 45 cm and area of cross-section `3 xx 10^(6) m^(2)`. The composite string is stretched by a tension of 80 N. Density of steel is `7800 kg m^(-3)` and that of aluminium is`2600 kg m^(-3)`. The minimum frequency of tuning fork which can produce standing wave in it with node at the joint is,

Options

A. 357.3 Hz
B. 375.3 Hz
C. 337.5 Hz
D. 325.5 Hz

Answer

C

Explanation

Using `v = n/(2l) sqrt(T/mu)` we get `v_("steel") = n_(1)/(2 xx 60 xx 10^(-2)) xx sqrt(80/(7800 xx 10^(-6)))`…………..(i)
`v_("aluminium") = n_(2)/(2 xx 45 xx 10^(-2)) sqrt(80/(3 xx 10^(-6) xx 2600))`…….(2)
Equating (1) of (2) `n_(1)/n_(2) = 0.6/0.45 = 4/3`
From (1) `v_("steel") = 4/(2 xx 0.6) sqrt(80/(780 xx 10^(-6))) = 337.5 Hz`
In composite string, different harmonic give rise to a common frequency.
`therefore v_("steel") = v_("aluminium") = 337 Hz`