Question

A block is connected with a spring executing SHM with time period $T$ in vertical plane. The amplitude of oscillation is $A .$ Find the time taken by the particle to move from position $y = + \frac{A}{2}$ to $y = - \frac{A}{2}$ in the following situation.

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Accepted Answer

Let us draw phasor circle corresponding to this situation as shown below.

The block moves from position $P,$ to $P_{2}$ on the phasor circle. From $P_{1}$ to $P_{2}$ the block turns through on angle $2 θ .$ It is also equal to change in phase of the block.

Hence, $\sin θ = \frac{1}{2}$ or $θ = \frac{π}{6}$

Hence, angle rotated, $2 θ = \frac{π}{3} = Δ ϕ$ (change in phase)

Thus, time taken for block in reaching from $P_{1}$ to $P_{2},$

$Δ t = \frac{Δ ϕ}{ω} = \frac{π}{3 ω}$

Here $ω = \frac{2 π}{T},$ hence $Δ t = \frac{π}{3 \times (2 π / T)} = \frac{T}{6}$

$Δ t = \frac{T}{2 π} \times Δ ϕ = \frac{T}{2 π} \times \frac{π}{3} = \frac{T}{6}$

Situation II: We can draw the phasor circle corresponding to this situation as shown below.

We can observe that the change in phase during the motion of the block from $P,$ to $P_{2}$ is $π$ or $Δ ϕ = π .$

Hence, time taken by the block, $Δ t = \frac{T}{2 π} \times Δ ϕ = \frac{T}{2 π} \times π = \frac{T}{2}$

B M Sharma Solutions for Chapter: Linear and Angular Simple Harmonic Motion, Exercise 1: ILLUSTRATIONS

B M Sharma Physics Solutions for Exercise - B M Sharma Solutions for Chapter: Linear and Angular Simple Harmonic Motion, Exercise 1: ILLUSTRATIONS

Questions from B M Sharma Solutions for Chapter: Linear and Angular Simple Harmonic Motion, Exercise 1: ILLUSTRATIONS with Hints & Solutions

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