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A mass is suspended separately by two sp...

A mass is suspended separately by two springs of spring constants `k_(1)` and `k_(2)` in successive order. The time periods of oscillations in the two cases are `T_(1)` and `T_(2)` respectively. If the same mass be suspended by connecting the two springs in parallel, (as shown in figure) then the time period of oscillations is T. The correct relations is

A

`T^(2)=T_(1)^(2)+T_(2)^(2)`

B

`T^(-2)=T_(1)^(-2)+T_(2)^(-2)`

C

`T^(-1)=T_(1)^(-1)+T_(2)^(-1)`

D

`T=T_(1)+T_(2)`

Text Solution

Verified by Experts

The correct Answer is:
B
(b) `T=2pisqrt((m)/(k)) " or " T prop (1)/(sqrt(k)) " or "k=(alpha)/(T^(2))`
Now , `k_(e)=k_(1)+k_(2)`
`therefore (alpha)/(T^(2))=(alpha)/(T_(1)^(2))+(alpha)/(T_(2)^(2))`
or `T^(-2)=T_(1)^(-2)+T_(2)^(-2)`
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