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Three simple harmonic motions in the same direction having the same amplitude a and same period are superposed. If each differs in phase from the next by \( 45^{\circ}, \) then : This question has multiple correct options

1. The resultant amplitude is \( (1+\sqrt{2}) a \)
2. The phase of the resultant motion relative to the first is \( 90^{\circ} \)
3. The energy associated with the resulting motion is \( (3+2 \sqrt{2}) \) times the energy associated with any single motion
4. The resulting motion is not simple harmonic

Correct answer: The energy associated with the resulting motion is \( (3+2 \sqrt{2}) \) times the energy associated with any single motion

Solution

For equal-frequency SHMs, the resultant is still SHM, with amplitude equal to the phasor sum. Here the three phasors at 0°, 45°, and 90° add to amplitude \((1+\sqrt2)a\), so energy scales as amplitude squared: \((1+\sqrt2)^2=3+2\sqrt2\).

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