1.

The maximum speed of a particle executing an S.H.M. is 1" m s"^(-1) and maximum acceleration is 1.57" m s"^(-2). The timeperiod of S.H.M. is :

Answer»

0.25 s
4.00 s
1.57 s
`(1)/(1.57)s.`

Solution :`v_("MAX")=r omega` and `a_("max")=omega^(2)r`.
`(a_("max"))/(v_("max"))=omega=(2PI)/(T)`
`T=2pi""(v_("max"))/(a_("max"))=2pi.(1)/(157)=4` s.
Hence correctchoice is (b).


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