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  • Consider a pair of identical pendulums, which oscillate with equal amplitude independently such that when one pendulum is at its extreme position making an angle of 2° to the right with the vertical, the other pendulum makes an angle of 1° to the left

Consider a pair of identical pendulums, which oscillate with equal amplitude independently such that when one pendulum is at its extreme position making an angle of 2^{\circ} to the right with the vertical, the other pendulum makes an angle of 1^{\circ} to the left of the vertical. What is the phase difference between the pendulums?


 

Answers (1)

Now, \Theta_{1} = \Theta_{0} \sin (\omega t + \delta_{1})

& \Theta_{2} = \Theta_{0} \sin (\omega t + \delta_{2})

Now, for the first pendulum,

\Theta = 2^{\circ}

Thus, \sin (\omega t + \delta _{1}) = 1

& for second pendulum,

\Theta = -1^{\circ}

Thus, \sin (\omega t + \delta _{2}) = -1

Thus, (\omega t + \delta _{1}) = 90^{\circ} & (\omega t + \delta _{2}) = 30^{\circ}

Therefore,

\delta _{1}+\delta _{2}=120^{\circ}

Posted by

infoexpert24

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