Question

A particle is executing SHM of amplitude $A,$ about the mean position $x=0.$ Which of the following is a possible phase difference between the positions of the particle at $x=+\frac{A}{2}$ and $x=-\frac{A}{\sqrt{2}}$.

• A. ${75}^{\circ }$
• B. ${165}^{\circ }$
• C. ${135}^{\circ }$
• D. ${195}^{\circ }$

Answer 1

Answer: (B), (D)

${165}^{\circ }$
${195}^{\circ }$

The equation for SHM $x=A\cos (\omega t+\varphi )$
When $x=\frac{A}{2},\Rightarrow \frac{A}{2}=A\cos (\omega t+{\varphi }_1)$
or ${60}^o=\omega t+{\varphi }_1....\left(1\right)$
When $x=-\frac{A}{\sqrt{2}},\Rightarrow -\frac{A}{\sqrt{2}}=A\cos (\omega t+{\varphi }_1)$
or ${225}^o=\omega t+{\varphi }_2....\left(2\right)$
$\left(2\right)-\left(1\right)\Rightarrow {\varphi }_2-{\varphi }_1=225-60={165}^o$
also $x=A\sin (\omega t+\varphi )$
When $x=\frac{A}{2},\Rightarrow \frac{A}{2}=A\sin (\omega t+{\varphi }_1)$
or ${30}^o=\omega t+{\varphi }_1....\left(1\right)$
When $x=-\frac{A}{\sqrt{2}},\Rightarrow -\frac{A}{\sqrt{2}}=A\sin (\omega t+{\varphi }_1)$
or ${225}^o=\omega t+{\varphi }_2....\left(2\right)$
$\left(2\right)-\left(1\right)\Rightarrow {\varphi }_2-{\varphi }_1=225-30={195}^o$