A wire of length l and mass per unit length 6.0×10−3 kg/m is put under tension of 540N two consecutive frequencies that it resonates at are 420 Hz and 490 Hz , then l in meters is _____
nv2l=420 → i n+1v2l=490 → (ii) Divide i/ii n=6 Substitute in equation (i) we have ⇒6×v2l=420 Also, v=Tμ=5406×10−3 ∴6×5406×10−32l=420 l=300140=2.142