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Two masses m1 and m2 are suspended toget...

Two masses m1 and m2 are suspended together by a massless spring of constant k. When the masses are in equilibrium, m1 is removed without disturbing the system. The amplitude of oscillations is

A

`(m_(1)g)/(K)`

B

`(m_(2)g)/(K)`

C

`((m_(1)+m_(2))g)/(K)`

D

`((m_(1)-m_(2))g)/(K)`

Text Solution

Verified by Experts

The correct Answer is:
A
`m_(2)g=kl` . . . (i)
l is the extension in the string when only `m_(2)` is connected.
Now, `(m_(1)+m_(2))g=k(l+Deltal)` . . . (ii)
`Deltal to ` increase in extension when `(m_(1)+m_(2))` is connected
from equation (i) and (ii)
`m_(1)g=k Deltal implies Deltal=(m_(1)g)/(k)`.
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