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Two masses m(1) and m(2) are suspended ...

Two masses `m_(1)` and `m_(2)` are suspended w together by a light spring of spring constant k as shown in figure. When the system is in equilibrium, the mass `m_(1)` is removed without disturbing the system. As a result of this removal, mass `m_(2)` performs simple harmonic motion. For this situation mark the correct statement(s).

A

The amplitude of osciallation is `(m_(1)g)/k`

B

The amplitude of oscillations is `((m_(1) + m_(2))g)/k`

C

The system oscillates with angular frequency `sqrt(k/m_(2))`

D

The system oscillates with angular frequency `sqrt(k/(m_(1)+m_(2)))`

Text Solution

Verified by Experts

The correct Answer is:
A, C
When both the blocks are connected then in equilibrium position, the elongation in spring is given by:
`y_(0) =((m_(1) + m_(2))g)/k`

In equilibrium position, for resulting oscillation after removal of `m_(1)`, elongation in spring is:
`y_(1) =(m_(2)g)/k`
So, amplitude of oscillations is,
`y_(0)-y_(1) =(m_(1)g)/k`
Time period of simple harmonic motion is,
`T =2pi sqrt(m_(2)/k) rArr omega = sqrt(k/m_(2))`
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