The equation of the resulting oscillation obtained by the summation of two mutually perpendicular oscillations with the same frequency f1 = f2 = 5 Hz and same initial phase δ1 = δ2 = 600 is, (Given, their amplitudes are A1 = 0.1 m and A2 = 0.05 m

When x = A1sin(ωt + ∅1) and y = A2 sin(ωt+∅2) thenx2A12+y2A22-2xyA1A2cosδ =  sin2δHere phase difference i.e., δ = 0∴  x2A21+y2A22-2xyA1A2 = 0⇒(xA1-yA2)2 = 0 ⇒y = A2A1xResultant amplitude isA = A21+A22 = (0.1)2+(0.05)2 = 0.112 mω = 2πf1 = 2πf2 = 10πThus the equation of resulting SHM isr = 0.112 sin(10πt+π3)