Consider a pair of identical pendulums,which | vedclass

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(A) Let the angular displacement of the two pendulums be given by:$	heta_1 = 	heta_0 \sin(\omega t + \phi_1)$$	heta_2 = 	heta_0 \sin(\omega t + \p

Vedclass · Accepted Answer

$2\pi/3$

Vedclass · Answer

(A) Let the angular displacement of the two pendulums be given by:$	heta_1 = 	heta_0 \sin(\omega t + \phi_1)$$	heta_2 = 	heta_0 \sin(\omega t + \phi_2)$where $	heta_0$ is the amplitude. Since the pendulums are identical and oscillate with equal amplitude,$	heta_0 = 2^o$.For the first pendulum at its extreme right position:$	heta_1 = +	heta_0 = 2^o$$2^o = 2^o \sin(\omega t + \phi_1) \implies \sin(\omega t + \phi_1) = 1 \implies \omega t + \phi_1 = \pi/2$For the second pendulum at $1^o$ to the left:$	heta_2 = -1^o = -	heta_0/2 = -2^o/2$$-2^o/2 = 2^o \sin(\omega t + \phi_2) \implies \sin(\omega t + \phi_2) = -1/2$Thus,$\omega t + \phi_2 = -\pi/6$ or $7\pi/6$.Taking $\omega t + \phi_2 = -\pi/6$,the phase difference $\Delta \phi = |(\omega t + \phi_1) - (\omega t + \phi_2)| = |\pi/2 - (-\pi/6)| = |\pi/2 + \pi/6| = |3\pi/6 + \pi/6| = 4\pi/6 = 2\pi/3$.

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