Question

Two particles are in SHM with same angular frequency and amplitudes A and 2A respectively along same straight line with same mean position. They cross each other at position distance from mean position in opposite direction. The phase difference between them is

• A. $\left( \frac { 1 } { 4 } \right)$
• B. $\frac { \pi } { 6 }$ – sin^–1$\left( \frac { 1 } { 4 } \right)$
• C. $\frac { 5 \pi } { 6 }$ – cos^–1 $\left( \frac { 1 } { 4 } \right)$
• D. $\frac { \pi } { 6 }$ – cos^–1 $\left( \frac { 1 } { 4 } \right)$

Answer 1

Answer: (A)

$\left( \frac { 1 } { 4 } \right)$

Answer 2

sin f₁ = $\frac { \mathrm { A } / 2 } { \mathrm {~A} }$ = $\frac { 1 } { 2 }$

f₁ = $\frac { \pi } { 6 }$

sin (p – f₂) = = $\frac { 1 } { 4 }$

f₂ = p – sin^–1 $\left( \frac { 1 } { 4 } \right)$

Phase difference

f₂ – f₁ = $\frac { 5 \pi } { 6 }$ – sin^–1 $\left( \frac { 1 } { 4 } \right)$