Two particles executing SHM possess the same frequency. When the first particle just passes the mean position of its path, the second particle moving in the same direction is at a distance of 3 cm from its mean position. If the amplitude of vibration of the second particle is 6 cm, when is the phase difference of the two particles?

Two particles executing SHM possess the same frequency. When the first

1.	Two particles executing SHM possess the same frequency. When the first particle just passes the mean position of its path, the second particle moving in the same direction is at a distance of 3 cm from its mean position. If the amplitude of vibration of the second particle is 6 cm, when is the phase difference of the two particles?
Answer» SOLUTION :If A is the amplitude and `theta` is the phase angle, then displacement, `x=Acostheta` For the first particle, `x_(1)=A_(1)costheta_(1)or,0=A_(1)costheta_(1)` or, `costheta_(1)=0or,theta_(1)=pm90^(@)` For the second particle, `x_(2)=A_(2)costheta_(2)or,costheta_(2)=x_(2)/A_(2)=3/6=1/2` or, `theta_(2)=pm60^(@)` At the given instant, the TWO particles are in motion in the ssame direction. So, (i) `theta_(1)=90^(@),"then "theta_(2)=60^(@),` phase difference = `theta_(2)-theta_(1)=-30^(@)` or, (II) If `theta_(1)=-90^(@),"then "theta_(2)=-60^(@),` phase difference `theta_(2)-theta_(1)=+30^(@)` `therefore" "` The required phase difference is `pm30^(@)`.