Two particles execute SHM with amplitude $A$ and $2 A$ and angular frequency $ω$ and $2 ω$ respectively. At $t = 0$ they starts with some initial phase difference. At $t = \frac{2 π}{3 ω}$ they are in same phase. Their initial phase difference is

Question

Two particles execute SHM with amplitude

A

and

2 A

and angular frequency

ω

and

2 ω

respectively. At

t = 0

they starts with some initial phase difference. At

t = \frac{2 π}{3 ω}

they are in same phase. Their initial phase difference is

Toppr Admin · Accepted Answer

Let initial phase of first particle be -x3B8-1-Its phase at time t is -x3B8-1-x3C9-t-x3B8-1-2-x3C0-3Let initial phase of first particle be -x3B8-2-Its phase at time t is -x3B8-2-2-x3C9-t-x3B8-2-4-x3C0-3As two phases are equal at time t-so-xA0-t is -x3B8-1-2-x3C0-3-x3B8-2-4-x3C0-3-x21D2-x3B8-1-x2212-x3B8-2-2-x3C0-3- is the phase difference-

Two particles execute SHM with amplitude A and 2A and angular frequency ω and 2ω respectively. At t=0 they starts with some initial phase difference. At t=2π3ω they are in same phase. Their initial phase difference is2π3ππ34π3

Let initial phase of first particle be θ1.Its phase at time t is θ1+ωt=θ1+2π/3Let initial phase of first particle be θ2.Its phase at time t is θ2+2ωt=θ2+4π/3As two phases are equal at time t,so, t is θ1+2π/3=θ2+4π/3⇒θ1−θ2=2π/3, is the phase difference.