Question

If the second pendulum bob is thrown at velocity $v_{2}$ at an angle $β$ from mean position, but on other side of mean position, find the phase difference in the two SHMs now as shown in the figure.

I

II

Accepted Answer

In this case, amplitudes of the two SHMs will still remain the same and are given by Eqs. (i) and (ii) but when we represent the two SHMs on their corresponding circular motions, the position of the particle in circular motion in second pendulum is now different as shown in the figure.

a

b

As shown in Figs. (a) and (b) the initial phases of the two pendulum bobs are

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

and

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

As angular frequency $ω$ for both SHM are same, their phase difference remains constant so it is given as

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

This is the phase difference between two simple harmonic motions.

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

Learn and Prepare for any exam you want !