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The system is in equilibrium and at rest...

The system is in equilibrium and at rest. Now mass `m_(1)` is removed from `m_(2)`. Find the time period and amplituded of resultant motion. Spring constant is `K`.

Text Solution

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Initial extension in the spring
`x = ((m_(1) - m_(2)g)/(K))`
Now, if we remove `m_(1)`, equilibrium position `(E.P)` of `m_(2)` will be `(m_(2)g)/(K)` below natureal length of spring

As the initial positon, since velocity is zero i.e., it is the extreme position.
Hence Amplitude
`= (m_(1)g)/(K)`
Time period `= 2pisqrt((m_(2))/(K))`
Since only block mass `m_(2)` is oscillating
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