Two particles are performing simple harmonic motion in a straight line about the same equilibrium point. The amplitude and time period for both particles are same and equal to A and T, respectively. At time t = 0 one particle has displacement A while the other one has displacement (-A/2) and they are moving towards each other. If they cross each other at time t, then t is :

Question

Two particles are performing simple harmonic motion in a straight line about the same equilibrium point. The amplitude and time period for both particles are same and equal to A and T, respectively. At time t = 0 one particle has displacement A while the other one has displacement (-A/2) and they are moving towards each other. If they cross each other at time t, then t is : (A) (T/6) (B) ( 5T/6) (C) (T/3) (D) (T/4)

Tardigrade Admin · Accepted Answer

angle covered to meet =60°=(π/3) rad t=(Q/w) =(π/3×2π) T=(T/6)

$\frac{T}{6}$

$\frac{5 T}{6}$

$\frac{T}{3}$

$\frac{T}{4}$

angle covered to meet $= 6 0^{\circ} = \frac{π}{3}$ rad
$t = \frac{Q}{w}$
$= \frac{π}{3 \times 2 π} T = \frac{T}{6}$

6T​

65T​

3T​

4T​

angle covered to meet =60∘=3π​ rad t=wQ​ =3×2ππ​T=6T​

$\frac{T}{6}$

$\frac{5 T}{6}$

$\frac{T}{3}$

$\frac{T}{4}$

angle covered to meet $= 6 0^{\circ} = \frac{π}{3}$ rad
$t = \frac{Q}{w}$
$= \frac{π}{3 \times 2 π} T = \frac{T}{6}$