Answer:
Angular
displacement of a pendulum at any instant is given by
\[\phi
={{\phi }_{0}}\,\,\sin \,(\omega t+\,\theta )\]
where,
\[{{\phi }_{0}}\] is maximum angular
displacement and \[\theta \]
is the
initial phase.
\[\therefore
\] For first pendulum,
\[{{2}^{o}}={{2}^{o}}\,\sin \,(\omega t+{{\theta }_{1}})\]
or \[\sin \,(\omega
t+{{\theta }_{1}})=1\] or \[\omega
t+{{\theta }_{1}}={{90}^{o}}\]?.. (i)
For
second pendulum, \[-{{1}^{o}}={{2}^{o}}\sin (\omega t+{{\theta }_{2}})\]
or \[\sin
\,(\omega t+{{\theta }_{2}})=-\frac{1}{2}\] or
\[(\omega t+{{\theta }_{2}})\]
\[=-{{30}^{o}}\] ?.
(ii)
From
eqns. (i) and (ii),
\[\omega
t+\,{{\theta }_{1}}\,-\omega t\,-{{\theta }_{1}}\,={{120}^{o}}\]
or \[({{\theta
}_{1}}+\,{{\theta }_{2}})\,={{120}^{o}}\]
You need to login to perform this action.
You will be redirected in
3 sec