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The system is in equilibrium and at rest...

The system is in equilibrium and at rest. Now mass m1 is removed from m2 . Find the time period and amplitude of resultant motion. (Given : 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 m1 . equillibrium position (E.P.) of m will be `(m_(2)g)/(K)`below natural length of spring. At the initial position, since velocity is zero i.e. it is the extreme position.
Hence Amplitude `= (m_(1)g)/(K)`
Time period `= 2pi sqrt(m_(2)/(K))`
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