Paragraph : A particle vibrates in S.H.M. along a straight line. Its velocity is 4 cm/s when its displacement is 3 cm and velocity is 3 cm/s when displacement is 4 cm. The length of a simple pendulum executing simple harmonic motion is increased by 21%. The percentage increase in the time period of the pendulum of increased length is :

Paragraph : A particle vibrates in S.H.M. along a straight line. Its v

1.	Paragraph : A particle vibrates in S.H.M. along a straight line. Its velocity is 4 cm/s when its displacement is 3 cm and velocity is 3 cm/s when displacement is 4 cm. The length of a simple pendulum executing simple harmonic motion is increased by 21%. The percentage increase in the time period of the pendulum of increased length is :
Answer» 0.11 0.21 0.42 0.1 Solution :`T=2pi SQRT((l)/(g))` and `T.=2pi sqrt((1.21l)/(g))` `(T.)/(T)=sqrt(1.21)implies(T.)/(T)=1.1` `implies""T.=1.1T` `:. DELTA T=0.1" T"` %increase in time PERIOD `=(DELTAT)/(T)xx100=(0.1T)/(T)xx100=10%` Hence correct choice is (d).

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