A) period
B) acceleration
C) mass
D) velocity
Correct Answer: B
Solution :
As we know that, \[\alpha ={{\omega }^{2}}x\] (in magnitude) \[\Rightarrow \] \[\alpha ={{\left( \frac{2\pi }{T} \right)}^{2}}x\] \[\Rightarrow \] \[T=2\pi \,\sqrt{\frac{x}{a}}\] \[\varepsilon =\sqrt{\frac{displacement}{acceleration}}\] and T in SUM is constant Therefore, ratio (displacement/acceleration) is constant.You need to login to perform this action.
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