Two masses `m_(1)=1.0 kg and m_(2)=0.5 kg ` are suspended together by a massless spring of force constant, `k=12.5 Nm^(-1)`. When they are in equillibrium position, `m_(1)` is gently removed. Calculate the angular frequency and the amplitude of oscillation of `m_(2)`. Given `g=10 ms^(-2)`.

Explanation

the given arrangement is shown in the below figure.
case I : When both blocks are suspended together
Let `y_(1)` be the extension produced in the case.
then, `F = (1.5 + 0.5)g = ky_(1)`
`rArr y_(1) (2g)/(k) = ( 2 xx 10)/(12) = 1.66` m
Case II : when only one block of mass 1.5 kg is supended
Let `y_(2)` be the extension produced in this case, Then,
1.5 g = `ky_(2)`
`y_(2) = (1.5)/(k) = (1.5 xx 10)/(12)`
= 1.25 m
`therefore y_(1) = y_(2) = 1.66 - 1.25 = 0.41`m
This will be amplitude of oscillation of block of mass 1.5 kg
A = 0.41 m
Angular frequency `omega = sqrt((k)/(1.5)) = sqrt((12)/(1.5)) = 2.83` rad/s