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A simple pendulum has a time period T(1)...

A simple pendulum has a time period `T_(1)`. When its point of suspension is moved vertically upwards according as `y=kt^(2)`, where y is vertical covered and `k=1 ms^(-2)` , its time period becomes `T_(2)` then ` (T_(1)^(2))/(T_(2)^(2))` is (g`=10 ms^(-2))`

A

`(5)/(6)`

B

`(11)/(10)`

C

`(6)/(5)`

D

`(4)/(5)`

Text Solution

Verified by Experts

The correct Answer is:
C
`y=kr^(2)`
`(dy)/(dt)=2kt,(d^(2)y)/(dt^(2))=2k`
Here `k=1ms^(-2)`
`(d^(2)y)/(dt^(2))=2xx1=2ms^(-2),g_(2)=g_(1)+2=10+2=12ms^(-2)`
For a pendulum, Time period `T=2pisqrt((l)/(g))`
`((T_(1))/(T_(2)))^(2)=(g_(2))/(g_(1))=(12)/(10)=(6)/(5)`
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