The equation of the resulting oscillation obtained by the summation at two mutually perpendicular oscillation with the same frequency `f_(1) = f_(2) = 5 Hz` and same initial phase `delta_(1) = delta_(2) = 60^(@) ` is (Given their amplitude are `A_(1) = 0.1m and A_(2) = 0.05m`

Options

A. `0.15sin(10pi t +(pi)/(6))`
B. `0.05sin(10pi t +(2pi)/(3))`
C. `0.112 sin(10pi t +(pi)/(3))`
D. `0.313 sin(10pi t +(pi)/(2))`

Answer

C

Explanation

When `x = A_(1) sin (omega t + phi_(1))`
and `y = A_(2) sin (omegat + phi_(2))` then
`(x^(2))/(A_(1)^(2)) + (y^(2))/(A_(2)^(2)) - (2xy)/(A_(1) A_(2)) cos delta = sin ^(2) delta`
Here phase i.e. `delta = 0`
`:.(x^(2))/(A_(1)^(2)) + (y^(2))/(A_(2)^(2)) - (2xy)/(A_(1) A_(2)) = 0`
`rArr ((x)/(A_(1)) - (y)/(A_(2)))^(2) = 0 rArr y = (A_(2))/(A_(1)) x`
REsulting amplitude is
`A = sqrt(A_(1)^(2) + A_(2)^(2)) = sqrt((0.1)^(2) + (0.05)^(2)) = 0.112m`
Thus the equation of resulting `SHM`
` r = 0.112 sin (10pi t + (pi)/(3))`