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A mass m is suspended separately by two ...

A mass `m` is suspended separately by two different springs of spring constant `k_(1)` and `k_(2)` given the time period `t_(1)` and `t_(2)` respectively. If the same mass `m` is shown in the figure then time period `t` is given by the relation

A

`t = t_(1) + t_(2)`

B

`t =( t_(1)t_(2))/(t_(1) + t_(2))`

C

`t^(2) = t_(1)^(2) + t_(2)^(2)`

D

`t^(-2) = t_(1)^(-2) + t_(2)^(-2)`

Text Solution

Verified by Experts

The correct Answer is:
D
`t_(1) = 2pi sqrt((m)/(k_(1)))` and `t_(2) = 2pi sqrt((m)/(k_(2)))`
Equivalant constant for shown combination is , `(K_(1) +K_(2)` so time period `t` is given by
`t= 2pi sqrt((m)/(K_(1) + K_(2)))`
By solving these equations we get `T^(2) = t_(1)^(2) + t_(2)^(2)`
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Knowledge Check

  • A mass is suspended separately by two different springs in successive order then time periods is t_(1) and t_(2) respectively. If it is connected by both spring as shown in figure then time period is to, the correct relation is

    A
    `t_(0)^(2) = t_(1)^(2) + t_(2)^(2)`
    B
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    A
    `T^(2) = T_(1)^(2) + T_(2)^(2)`
    B
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    D
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    B
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    C
    `sqrt((t_(1)t_(2))/(t_(1)+t_(2)))`
    D
    `t_(1)+t_(2)`

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