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Two springs of force constants 300 N/m (...

Two springs of force constants `300 N/m` (Spring A) and `400 N/m` (Spring B) are joined together in series. The combination is compressed by `8.75 cm.` The ratio of energy stored in A and B is `(E_(A))/(E_(B)) "Then" (E_(A))/(E_(B))` is equal to:

A

`(4)/(3)`

B

`(16)/(9)`

C

`(3)/(4)`

D

`(9)/(16)`

Text Solution

Verified by Experts

The correct Answer is:
A
Given :`K_(A)300 N/m,K_(B)=400 N/m`
Let when the combination of springs is compressed by force. F spring A is compressed by x.Therefore compression in spring B
`x_(B)=(8.75-x)`cm
F=`300xx x=400(8.75-x)`
Solving we get ,x=5 cm `x_(B)`=8.75-5=3.75 cm
`(E_(A))/(E_(B))=((1)/(2)K_(A)(x_(A))^(2))/((1)/(2)K_(B)(x_(B))^(2))=(300xx(5)^(2))/(400xx(3.75))=(4)/(3)`
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