Two particles P1 and P2 are executing SHM along the same straight line, whose equations are given x1=Asinωt+δ1 and x2=Asinωt+δ2. An observer on the ground, at t=0, observes particle P1 at distance A2 moving to the right from mean position O while particle P2 at -32 A moving to the left from mean position O, as shown in figure. Find the value of δ2-δ1.

x1=Asinωt+δ1At t=0, x=A2A2=Asinω(0)+δ1 sinδ1=12⇒δ1=π4 x2=Asinωt+δ2 x=-32A⇒ -32A=Asinω(0)+δ2 sinδ2=-32 ⇒ δ2=π+π3=4π3 δ2-δ1=4π3-π4=16-312π=1312π

Two particles P1 and P2 are executing SHM along the same straight line, whose equations are given x1=Asinωt+δ1 and x2=Asinωt+δ2. An observer on the ground, at t=0, observes particle P1 at distance A2 moving to the right from mean position O while particle P2 at -32 A moving to the left from mean position O, as shown in figure. Find the value of δ2-δ1.

x1=Asinωt+δ1At t=0, x=A2A2=Asinω(0)+δ1 sinδ1=12⇒δ1=π4 x2=Asinωt+δ2 x=-32A⇒ -32A=Asinω(0)+δ2 sinδ2=-32 ⇒ δ2=π+π3=4π3 δ2-δ1=4π3-π4=16-312π=1312π

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Two particles $P_{1}$ and $P_{2}$ are executing SHM along the same straight line, whose equations are given $x_{1} = A \sin (ω t + δ_{1})$ and $x_{2} = A \sin (ω t + δ_{2})$ . An observer on the ground, at t=0, observes particle $P_{1}$ at distance $\frac{A}{\sqrt{2}}$ moving to the right from mean position O while particle $P_{2}$ at $- \frac{\sqrt{3}}{2}$ A moving to the left from mean position O, as shown in figure. Find the value of $δ_{2} - δ_{1}$ .

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$\frac{7 π}{12}$

$\frac{13 π}{12}$

$\frac{5 π}{6}$

$π$

answer is B.

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Detailed Solution

$x_{1} = A \sin (ω t + δ_{1})$

At t=0, $x = \frac{A}{\sqrt{2}}$

$\frac{A}{\sqrt{2}} = A \sin [ω (0) + δ_{1}] \sin δ_{1} = \frac{1}{\sqrt{2}} \Rightarrow δ_{1} = \frac{π}{4} x_{2} = A \sin (ω t + δ_{2}) x = - \frac{\sqrt{3}}{2} A$

$\Rightarrow - \frac{\sqrt{3}}{2} A = A \sin [ω (0) + δ_{2}] \sin δ_{2} = - \frac{\sqrt{3}}{2} \Rightarrow δ_{2} = π + \frac{π}{3} = \frac{4 π}{3} δ_{2} - δ_{1} = \frac{4 π}{3} - \frac{π}{4} = (\frac{16 - 3}{12}) π = \frac{13}{12} π$