Two particles are oscillating in SHM along tw | vedclass

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(C) The distance between the two particles at any time $t$ is given by $d = |x_2 - x_1| = |A \sin(\omega t + \phi) - A \sin \omega t|$.Using the trigo

Vedclass · Accepted Answer

$74$

Vedclass · Answer

(C) The distance between the two particles at any time $t$ is given by $d = |x_2 - x_1| = |A \sin(\omega t + \phi) - A \sin \omega t|$.Using the trigonometric identity $\sin C - \sin D = 2 \sin(\frac{C-D}{2}) \cos(\frac{C+D}{2})$,we get:$d = |2A \sin(\frac{\phi}{2}) \cos(\omega t + \frac{\phi}{2})|$.The maximum distance occurs when the magnitude of the cosine term is $1$,so $d_{max} = 2A \sin(\frac{\phi}{2})$.Given $d_{max} = \frac{6A}{5}$,we have $2A \sin(\frac{\phi}{2}) = \frac{6A}{5}$.$\sin(\frac{\phi}{2}) = \frac{3}{5}$.Since $\sin(37^{\circ}) \approx 0.6 = \frac{3}{5}$,we have $\frac{\phi}{2} = 37^{\circ}$.Therefore,$\phi = 74^{\circ}$.

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