Two simple harmonic motions are given by x1=a2sin⁡ωt+3a2cos⁡ωt and x2=asin⁡ωt+acos⁡ωt above, the minimum phase difference between the two simple harmonic motions is (in degrees)

Amplitude of the first simple harmonic motion isA1=a22+3a22=aAmplitude of the second motion isA2=a2+a2=2a∴ A1A2=12We can rewrite the two motions asx1=a12sin⁡ωt+32cos⁡ωt=acos⁡60∘sin⁡ωt+sin⁡60∘cos⁡ωtor  x1=asin⁡ωt+60∘   ........(1)and  x2=a212sin⁡ωt+12cos⁡ωt=a2cos⁡45∘sin⁡ωt+sin⁡45∘cos⁡ωtor  x2=a2sin⁡ωt+45∘ ........(2)From (1) and (2) it follows that the amplitudes are a and a2 (a result obtained above) the phases are 60 and 45°. Therefore, phase difference = 60°- 45° = 15°