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Two masses m(1)=1.0 kg and m(2)=0.5 ...

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)`.

Text Solution

Verified by Experts

The correct Answer is:
`0.8`
Let y be the extension in the length of the spring when both ` m_(1) and m_(2)` are suspended
Then, ` F = (m_(1) + m_(2)) g = Ky`
or ` y = ((m_(1) + m_(2))g)/(K)`
Let the extension be reduced to y. when `m_(1)` is removed then
`m_(2) g = Ky.`
or ` y. = (m_(2) g)/( K)`
`:. y - y. = ((m_(1) + m_(2))g)/(K) - (m_(2) g)/( K)`
This will be the amplitude of oscillation of `m_(2)`
`:.` Amplitude, ` A = (m_(1) g)/( K) = (1.0 xx 10)/( 12.5) = 0.8 m`
Angular frequency
`omega = sqrt((K)/( m_(2))) = sqrt(( 12.5)/(0.5)) = 5 rad s^(-1)`
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