Question

Three simple harmonic motions in the same direction having the same amplitude and same period are superimposed. If each differ in phase from the other by ${45}^{\circ }$, then:

• 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 phase of the resultant motion relative to the first is ${45}^{\circ }$.
• D. The resulting motion is not simple harmonic.

Answer 1

Answer: (A), (C)

The correct options are
A The resultant amplitude is $(1+\sqrt{2})a$.
C The phase of the resultant motion relative to the first is ${45}^{\circ }$.

From the data given in the question ,
$y_1=a\sin \omega t.....\left(1\right)$
$y_2=a\sin (\omega t+\frac{\pi }{4})....\left(2\right)$
$y_3=a\sin (\omega t+\frac{\pi }{2})....\left(3\right)$
From the principle of superposition ,
$y=y_1+y_2+y_3$
$y=a(\sin \omega t+\sin (\omega t+\frac{\pi }{4})+\sin (\omega t+\frac{\pi }{2}))$
By using , $\sin C+\sin D=2\sin \left(\frac{C+D}{2}\right)\cos \left(\frac{C-D}{2}\right)$
$y=2a\sin (\omega t+\frac{\pi }{4})\cos \frac{\pi }{4}+a\sin (\omega t+\frac{\pi }{4})$
$\Rightarrow$ Resultant $y=(\sqrt{2}+1)a\sin (\omega t+\frac{\pi }{4})......\left(4\right)$
This is an equation of SHM.
So, Resultant amplitude, $A=(\sqrt{2}+1)a$,
Phase difference between $\left(1\right)$ and $\left(4\right)$ is $\frac{\pi }{4}$.
Hence, (a) & (c) are correct answers.