1.

A particle vibrating simple harmonically has an acceleration of 16 cm s^(-2)when it is at a distance of 4 cm from the mean position. Its time period is

Answer»

1s
2.571 s
3.142 s
6.028 s

Solution :Here, `a = 16 CM s^(-2), x= 4 cm, T=?`
We know, `|a|=omega^(2)x` or `omega^(2) =a/x`
`omega =(2PI)/T RARR 2 = (2pi)/T rArr T =(2pi)/2 = pi s = 3.142 s`


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