Three simple harmonic motions in the same dir | vedclass

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(D) Let the three simple harmonic motions be represented by:$y_1 = a \sin(\omega t - 45^\circ)$$y_2 = a \sin(\omega t)$$y_3 = a \sin(\omega t + 45^\ci

Vedclass · Accepted Answer

Both $(a)$ and $(c)$

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(D) Let the three simple harmonic motions be represented by:$y_1 = a \sin(\omega t - 45^\circ)$$y_2 = a \sin(\omega t)$$y_3 = a \sin(\omega t + 45^\circ)$On superimposing,the resultant $SHM$ is $y = y_1 + y_2 + y_3$$y = a[\sin(\omega t - 45^\circ) + \sin(\omega t) + \sin(\omega t + 45^\circ)]$Using the identity $\sin(A-B) + \sin(A+B) = 2\sin A \cos B$:$y = a[2\sin(\omega t)\cos(45^\circ) + \sin(\omega t)]$Since $\cos(45^\circ) = \frac{1}{\sqrt{2}}$:$y = a[2\sin(\omega t) \cdot \frac{1}{\sqrt{2}} + \sin(\omega t)]$$y = a[\sqrt{2}\sin(\omega t) + \sin(\omega t)] = a(1 + \sqrt{2})\sin(\omega t)$The resultant amplitude is $A = (1 + \sqrt{2})a$.Energy $E$ in $SHM$ is proportional to the square of the amplitude $(E \propto A^2)$:$\frac{E_{	ext{resultant}}}{E_{	ext{single}}} = \left(\frac{A}{a}ight)^2 = (1 + \sqrt{2})^2 = 1 + 2 + 2\sqrt{2} = (3 + 2\sqrt{2})$Thus,$E_{	ext{resultant}} = (3 + 2\sqrt{2})E_{	ext{single}}$.

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:

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