A particle is executing $SHM$ of amplitude $A,$ about the mean position $x = 0.$ Which of the following cannot be a possible phase difference between the positions of the particle at $x = +\,A/2$ and $x = - A/\sqrt {2} .$
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The equation for $\operatorname{SHM} x=A \cos (\omega t+\phi)$
When $x=\frac{A}{2}, \Rightarrow \frac{A}{2}=A \cos \left(\omega t+\phi_{1}\right)$
or $60^{\circ}=\omega t+\phi_{1} \ldots . .(1)$
When $x=-\frac{A}{\sqrt{2}}, \Rightarrow-\frac{A}{\sqrt{2}}=A \cos \left(\omega t+\phi_{1}\right)$
or $225^{\circ}=\omega t+\phi_{2} \ldots . .(2)$
$(2)-(1) \Rightarrow \phi_{2}-\phi_{1}=225-60=165^{\circ}$
also $x=A \sin (\omega t+\phi)$
When $x=\frac{A}{2}, \Rightarrow \frac{A}{2}=A \sin \left(\omega t+\phi_{1}\right)$
or $30^{\circ}=\omega t+\phi_{1} \ldots . .(1)$
When $x=-\frac{A}{\sqrt{2}}, \Rightarrow-\frac{A}{\sqrt{2}}=A \sin \left(\omega t+\phi_{1}\right)$
or $225^{\circ}=\omega t+\phi_{2} \ldots . .(2)$
$(2)-(1) \Rightarrow \phi_{2}-\phi_{1}=225-30=195^{\circ}$
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