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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 \( 45^{0}, \) then

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

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

Solution

For SHMs of the same frequency, the resultant is also SHM, with amplitude given by phasor addition. Since energy is proportional to amplitude squared, the energy ratio is the square of the resultant amplitude ratio. Here that ratio becomes \(3+2\sqrt{2}\).

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