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Two bodies A and B whose masses are in t...

Two bodies A and B whose masses are in the ratio `1:2` are suspended from two separate massless springs of force constants `k_(A) and k_(B)` respectively. If the two bodies oscillate vertically such that their maximum velocities are in the ratio `1:2` the ratio of the amplitude A to that of B is :

A

`sqrt((k_(B))/(2k_(A)))`

B

`sqrt((k_(B))/(8k_(A)))`

C

`sqrt((2k_(B))/(k_(A)))`

D

`sqrt((8k_(B))/(k_(A)))`

Text Solution

Verified by Experts

The correct Answer is:
B
The maximum velocity in the Oscillation is given as `v_(max)=Aomega`
The ratio of maximum velocity, `v_(A):v_(B)=1:2`
`2v_(A)=v_(B)`
`2A_(A)omega_(A)=A_(B)omega_(B)`
`(A_(A))/(A_(B))=(omega_(B))/(2omega_(B))` where, `omega=sqrt((k)/(m))`
`(A_(A))/(A_(B))=(1)/(2)sqrt((k_(B))/(2m_(A))xx(m_(B))/(k_(A)))=sqrt((1)/(4)xx(k_(B))/(2k_(A))) (m_(A))/(m_(B))=(1)/(2),2m_(A)=m_(B)`
`(A_(A))/(A_(B))=sqrt((K_(B))/(8K_(A)))`
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