First slide
Simple harmonic motion
Question

Two simple harmonic motions are given by x1=a2sinωt+3a2cosωt and x2=asinωt+acosωt above, the minimum phase difference between the two simple harmonic motions is (in degrees)

Moderate
Solution

Amplitude of the first simple harmonic motion is

A1=a22+3a22=a

Amplitude of the second motion is

A2=a2+a2=2a A1A2=12

We can rewrite the two motions as

x1=a12sinωt+32cosωt=acos60sinωt+sin60cosωtor  x1=asinωt+60   ........(1)and  x2=a212sinωt+12cosωt=a2cos45sinωt+sin45cosωtor  x2=a2sinωt+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°

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