Standing waves are produced by the superposition of
two waves
`y_(1)=0.05 sin (3 pit-2x) and y_(2) = 0.05 sin (3pit+2x)`
Where x and y are in metres and t is in second. What
is the amplitude of the particle at `x = 0.5` m ? (Given,
`cos 57. 3 ^(@) = 0. 54)`

Options

A. 2.7 cm
B. 5.4 cm
C. 8.1 cm
D. 10.8 cm

Answer

C

Explanation

Here, `y_(1)=0.05 sin(3pit-2x)`
`y_(2)=0.05 sin(3pit+2x)`
According to superposition principle, the resultant displacement is `y=y_(1)+y_(2)=0.05[sin(3pit-2x)+sin(3pit+2x)]`
or `y=0.05xx2sin3pitcos2x`
or `y=(0.1cos2x)sin3pit=Asin3pit`
where `A=0.1cos2x` = amplitude of the resultant standing wave At `x=0.5m`,
`A=0.1cos2x=0.1cos(2xx0.5)=0.1cos1` (radian)
`=0.1cos.(180^(@))/(pi)=0.1cos57.3^(@)`
or `A=0.1xx0.54m=0.054m=5.4m`