Question

Figure shows two identical simple pendulums of length $l .$ One is tilled at an angle $α$ and imparted an initial velocity $v_{1}$ toward mean position and at the same time other one is projected away from mean position at a velocity $v_{2}$ at an initial angular displacement $β .$ Find the phase difference in oscillations of these two pendulums.

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Accepted Answer

It is given that the first pendulum bob is given a velocity $v_{1}$ at a displacement $l α$ from the mean position, using the formula for the velocity we can find the amplitude of its oscillations as

$v_{2}$ It $A,$ is the amplitude of SHM of this bob.

As for simple pendulum $ω = \sqrt{\frac{g}{l}}$ we have

$v_{1}^{2} = \frac{g}{l} [(A_{1}^{2}) - (l α)^{2}]$

$A_{1} = \sqrt{l^{2} α^{2} + \frac{v_{1}^{2} l}{g}} \dots (1)$

Similarly, it $A_{2}$ is the amplitude of SHM of the second pendulum bob we have

$v_{2} = ω \sqrt{A_{2}^{2} - (l β)^{2}}$

or $A_{2} = \sqrt{l^{2} β^{2} + \frac{v_{2}^{2} l}{g}} \dots (2)$

Now we represent the two SHMs by circular motion representation as shown in the figure.

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In the first pendulum, $t = 0$ the bob is thrown from a displacement $l α$ from a mean position with a velocity $v_{1}$ towards the mean position. As it is moving towards the mean position, in Fig. (a), we consider the corresponding circular motion particle $A_{0}$ of the bob $A$ is the second quadrant, as the reference direction of $ω,$ we consider anticlockwise. As shown in the figure initial phase of bob $A$ is given as

$ϕ_{1} = π - θ$

here $\sin θ = \frac{l α}{A_{1}}$

$ϕ_{1} = π - \sin^{- 1} (\frac{l α}{A_{1}}) \dots (3)$

Similarly, for the second pendulum bob, $B,$ its corresponding circular motion particle $B_{0}$ a $t = 0$ is considered as shown in Fig. (b) its initial base is given as

$ϕ_{2} = \sin^{- 1} (\frac{l β}{A_{2}}) \dots (4)$

As both the pendulums are identical, their angular frequency for SHM must be the same, so their phase difference will not change with time, hence their phase difference can be given as

$Δ ϕ = ϕ_{2} - ϕ_{1} = \sin^{- 1} (\frac{l β}{A_{2}}) + \sin^{- 1} (\frac{l α}{A_{1}}) - π$

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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