A transverse wave propagations on a stretched string of linear density 3×10−4 kg/m is represented by equation y=0.2 Sin(1.5 x+30t) where 'x' is in meters and 't' is in seconds. The tension in string (in newtons) is
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a
12 N
b
12×10−3 N
c
12×10−2 N
d
1.2 N
answer is C.
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Detailed Solution
Given, Linear density μ=3×10−4 kg/m y=0.2 Sin(1.5 x+30t) ……(1)y=Asin(kx+ωt) ………….(2)Compare (1) and (2)A=0.2 ,K=1.5 ,ω=30 v=ωk=301.5 v=20 v=Tμ 20=T3×10−4 400=T3×10−4 T=12×10−2 N