Simple Harmonic Motion Ques 59
- Three simple harmonic motions in the same direction having the same amplitude and same period are superposed. If each differ in phase from the next by $45^{\circ}$, then
(1999, 3M)
(a) the resultant amplitude is $(1+\sqrt{2}) a$
(b) the phase of the resultant motion relative to the first is $90^{\circ}$
(c) the energy associated with the resulting motion is $(3+2 \sqrt{2)}$ times the energy associated with any single motion
(d) the resulting motion is not simple harmonic
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Answer:
Correct Answer: 59.(a,c)
Solution:
Formula:
- From superposition principle
$ \begin{aligned} y & =y _1+y _2+y _3 \\ & =a \sin \omega t+a \sin \left(\omega t+45^{\circ}\right)+a \sin \left(\omega t+90^{\circ}\right) \\ & =a\left[\sin \omega t+\sin \left(\omega t+90^{\circ}\right)\right]+a \sin \left(\omega t+45^{\circ}\right) \\ & =2 a \sin \left(\omega t+45^{\circ}\right) \cos 45^{\circ}+a \sin \left(\omega t+45^{\circ}\right) \\ & =(\sqrt{2}+1) a \sin \left(\omega t+45^{\circ}\right)=A \sin \left(\omega t+45^{\circ}\right) \end{aligned} $
Therefore, resultant motion is simple harmonic of amplitude
$ A=(\sqrt{2}+1) a $
and which differ in phase by $45^{\circ}$ relative to the first.
Energy in SHM $\propto(\text { amplitude })^{2}\left[E=\frac{1}{2} m A^{2} \omega^{2}\right]$
$ \begin{array}{ll} \therefore & \frac{E _{\text {resultant }}}{E _{\text {single }}}=(\frac{A}{a})^{2}=(\sqrt{2}+1)^{2}=(3+2 \sqrt{2}) \\ \therefore & E _{\text {resultant }}=(3+2 \sqrt{2}) E _{\text {single }} \end{array} $
BETA
