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

Question

Two particles are in SHM with same angular frequency and amplitudes A and 2 A respectively along same straight line with same mean position. They cross each other at position ( A /2) distance from mean position in opposite direction. The phase difference between them is: (A) (5 π/6)- sin -1((1/4)) (B) (π/6)- sin -1((1/4)) (C) (5 π/6)- cos -1((1/4)) (D) (π/6)- cos -1((1/4))

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/29625d0edd272939beb112a0dbc4c9a9-.png" /> sin φ1=( A / 2/ A )=(1/2) φ1=(π/6) sin (π-φ2)=( A / 2/2 A )=(1/4) φ2=π- sin -1((1/4)) Phase difference φ2-φ1=(5 π/6)- sin -1((1/4))

Q. Two particles are in $S H M$ with same angular frequency and amplitudes $A$ and $2 A$ respectively along same straight line with same mean position. They cross each other at position $\frac{A}{2}$ distance from mean position in opposite direction. The phase difference between them is:

$\frac{5 π}{6} - sin^{- 1} (\frac{1}{4})$

$\frac{π}{6} - sin^{- 1} (\frac{1}{4})$

$\frac{5 π}{6} - cos^{- 1} (\frac{1}{4})$

$\frac{π}{6} - cos^{- 1} (\frac{1}{4})$

$sin ϕ_{1} = \frac{A /2}{A} = \frac{1}{2}$
$ϕ_{1} = \frac{π}{6}$
$sin (π - ϕ_{2}) = \frac{A /2}{2 A} = \frac{1}{4}$
$ϕ_{2} = π - sin^{- 1} (\frac{1}{4})$
Phase difference
$ϕ_{2} - ϕ_{1} = \frac{5 π}{6} - sin^{- 1} (\frac{1}{4})$

Q. 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 2A​ distance from mean position in opposite direction. The phase difference between them is:

65π​−sin−1(41​)

6π​−sin−1(41​)

65π​−cos−1(41​)

6π​−cos−1(41​)