In the shown arrangement, both the spring are in their natural lengths. The coefficient of friction between `m_(2)` and `m_(1)` is `mu`. There is no friction between `m_(1)` and the surface. If the blocks are displaced slightly, they together perform simple harmonic motion. Obtain

(a) Frequency of such oscillations.
(b) The condition if the friction force on clock `m_(2)` is to act in the direction of its displacement from mean position.
( c) If the condition obtained in (b) is met, what can be maximum of their oscillations ?

Answer

`(1)/(2 pi) sqrt((K_(1) + K_(2))/(m_(1) + m_(2)))`

Explanation

a. For the oscillation of the blocks together, the equivalent, force constant equals `K_(1) + K_(2)` and hence the frequency
`A = (1)/(2 pi) sqrt((K_(1) + K_(2))/(m_(1) + m_(2)))`
b. Assuming that the block `(m_(1) + m_(2))` are displaced towards right, the various forces (except the frictional force) on `m_(2)`, in the reference frame of `m_(1)` are shown in the following free body diagram.

where `a = ((K_(1) + K_(2)x)/(m_(1) + m_(2)))`
For the frictional force to act on `m_(2)` in the direction of its displacement
`K_(2)x gt m_(2) [(K_(1) + K_(2))/(m_(1) + m_(2))]x`
`m_(1) K_(2) + m_(2) K_(2) - m_(2) K_(2) gt 0`
`m_(1) K_(2) - m_(2) K_(1) gt 0`
`(m_(1))/(m_(2)) gt (K_(1))/(K_(2))`
The above inequality is in the desired condition.
c.
Assuming `(m_(1))/(m_(2)) gt (K_(1))/(K_(2))` the FBD of `m_(2)` with `m_(1)` and` m_(2)`
together displacement towardsright can be shown as

If A be the amplitude of oscillation of `m_(2) K_(2) A - f = m_(2) omega^(2) A` (for the exterme position)
`f = K_(2) A - m_(2) omega^(2) A implies f = A (K_(2) - m_(2) omega^(2))`
`f le mu m_(2) g implies mu m_(2) g ge A (K_(2) - m_(2) omega^(2))`
`A le (mu m_(2) g)/((K_(2) - m_(2) omega^(2))) implies A_(max) = ((mu m_(2) g)/((K_(2) - m_(2) omega^(2))))`
Here `omega^(2) =m ((K_(1) + K_(2)))/((m_(1) + m_(2))) , A_(max) = ((mu m_(2) g (m_(1) + m_(2)))/(m_(1) K_(2) - m_(2) K_(1)))`