JEE Main 2016 · 9 Apr · Shift 1 · Physics Q10

← JEE Main 2016 · 9 Apr · Shift 1

PhysicsSingle correct+4 / -1Question 10 of 90

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 $\dfrac{-A}2$ and they are moving towards each other. If they cross each other at time t, then t is :

\dfrac T6

\dfrac{5T}6

\dfrac T3

\dfrac T4

Answer and worked solution

Official answer: A

Take $x_1=A\cos\omega t$ (starting at A, moving towards −x) and $x_2=A\cos(\omega t+\phi)$ with $\cos\phi=-\frac12$ and moving towards +x, so $\phi=\frac{4\pi}3$ . They meet when $\omega t=-\left(\omega t+\frac{4\pi}3\right)+2\pi$ , i.e. $\omega t=\frac\pi3$ : $t=\frac T6$ .

Topics:Simple harmonic motion

← Previous question Next question →

JEE Main 2016 · 9 Apr · Shift 1, Physics question 10