1.

Two simple pendulum of lengths 16I and I are in phase at the mean position at a certain time.IF T is the time period of shorter pendulum. The maximum time after which they will be again in phase.

Answer»

`1/3T`
`2/3T`
`3/2T`
`4/3T`

Solution :`T=2pisqrt(L//g`
`Talpha sqrt(l)`
The PENDULUM having time period makes one oscillation LESS than the other.
`(T_1-1)/(T_1)=sqrt(l)/(16l)=1/4`
`implies (T_1-1)/(T_1)=1/4`
`implies1-(1)/T_1=1/4`
`T=4//4`
Theywill be again in PHASE after`4/3T`,


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