Time period (T) and amplitude (A) are same for two particles which undergo SHM along the same line. At one particular instant, one particle is at phase (3 π/2) and other is at phase zero. While moving in the same direction. Find the time at which they will cross each other.

Question

Time period (T) and amplitude (A) are same for two particles which undergo SHM along the same line. At one particular instant, one particle is at phase (3 π/2) and other is at phase zero. While moving in the same direction. Find the time at which they will cross each other. (A) 4T/2 (B) 3T/8 (C) 3T/4 (D) 3T/7

Tardigrade Admin · Accepted Answer

At phase 3 π / 2,(ω t+φ)=3 π / 2 x1=A sin (3 π/2)=-A x2=0 So, at t=0, x1=-A cos ω t and x2=A sin ω t x1=x2 - cos ω t= sin ω t tan ω t=-1 ω t=(3 π/4) ((2 π/T)) t=(3 π/4) t=3 T / 8

Q. Time period $(T)$ and amplitude $(A)$ are same for two particles which undergo SHM along the same line. At one particular instant, one particle is at phase $\frac{3 π}{2}$ and other is at phase zero. While moving in the same direction. Find the time at which they will cross each other.

4T/2

3T/8

3T/4

3T/7

At phase $3 π /2, (ω t + ϕ) = 3 π /2$
$x_{1} = A sin \frac{3 π}{2} = - A$
$x_{2} = 0$
So, at $t = 0, x_{1} = - A cos ω t$ and $x_{2} = A sin ω t$
$x_{1} = x_{2}$
$- cos ω t = sin ω t$
$tan ω t = - 1$
$ω t = \frac{3 π}{4}$
$(\frac{2 π}{T}) t = \frac{3 π}{4}$
$t = 3 T /8$

Q. Time period (T) and amplitude (A) are same for two particles which undergo SHM along the same line. At one particular instant, one particle is at phase 23π​ and other is at phase zero. While moving in the same direction. Find the time at which they will cross each other.