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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^o, then

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The resultant amplitude is

\sqrt{2} a

The phase of the resultant motion relative to the first is 90^o

The energy associated with the resulting motion is

(3 + 2 \sqrt{2)}

times the energy associated with any single motion

The resulting motion is not simple harmonic

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detailed solution

Correct option is C

Let simple harmonic motions be represented by

$y_{1} = a \sin (ω t - \frac{π}{4}); y_{2} = a \sin ω t and y_{3} = a \sin (ω t + \frac{π}{4})$

$Resultant will be y = a [\sin (ω t - \frac{π}{4}) + \sin ω t + \sin (ω t + \frac{π}{4})]$

$\begin{array}{l} = a [2 \sin ω t \cos \frac{π}{4} + \sin ω t] \\ = a [\sqrt{2} \sin ω t + \sin ω t] = a (1 + \sqrt{2}) \sin ω t \end{array}$

Resultant amplitude $= (1 + \sqrt{2}) a$

Energy is S.H.M. $\propto (Amplitude)^{2}$

$\begin{array}{l} ∴ \frac{E_{Resultant}}{E_{Single}} = {(\frac{A}{a})}^{2} = (\sqrt{2} + 1)^{2} = (3 + 2 \sqrt{2}) \\ \Rightarrow E_{Resultant} = (3 + 2 \sqrt{2}) E_{Single} \end{array}$

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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 45o, then

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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^o, then