A) \[10\,m{{s}^{-2}}\]
B) \[7.5\,m{{s}^{-2}}\]
C) \[5\,m{{s}^{-2}}\]
D) \[5\sqrt{3}\,m{{s}^{-2}}\]
E) \[2.5\,m{{s}^{-2}}\]
Correct Answer: C
Solution :
For a simple pendulum time period \[T=2\pi \sqrt{\frac{l}{g}}\] or \[\frac{2\pi }{T}=\sqrt{\frac{g}{l}}\] \[\therefore \] \[\omega =\sqrt{\frac{g}{l}}\] \[{{\omega }^{2}}=\frac{g}{l}\] ?(i) Amplitude, when angular displacement is\[60{}^\circ \] \[=\frac{2\pi l}{360}\times 60\] \[=\frac{2\pi l}{6}\] Therefore, displacement when angular displacement is\[30{}^\circ \] \[=\frac{1}{2}\left( \frac{2\pi l}{6} \right)\] \[y=\frac{\pi l}{6}\] ?(ii) Acceleration \[(\alpha )=-{{\omega }^{2}}y\] Using Eq. (i) and (ii), we get \[\alpha =-\frac{g}{l}\times \frac{\pi l}{6}\] \[=-\frac{10\times 3.14}{6}\] \[=-5.2\,m{{s}^{-2}}\] \[\approx -5\,m{{s}^{-2}}\]
You need to login to perform this action.
You will be redirected in
3 sec
