A body A mass `m_(1) = 1 kg` and body B of mass `m_(2) = 4.1 kg`. The body A perform free vertical harmonic oscillations with the amplitude `1.6 cm` and frequency `25 Hz`. Neglecting the mass of the spring , find the maximum and m inimum value of force that the system exerts on the hearing surface.

Answer

`(m_(1) + m_(2)) g - m_(1) omega^(2) A`

Explanation

As A oscillates up and down , the normal reation berween B and the surface is maximum , where A is at position and it is minimum when A is at topmost position
Case I: Lowest position of A

Let `T_(1)` be the force in the compressed spring As B is at rest, `T_(1) - m_(2)g = R_(max)`
As acceleration of A is `omega^(2) A` towards mean position (up).
`T_(1) - m_(1)g = m_(1) omega^(2)`
Combining, ` R_(max) = (m_(1) + m_(2)) g + m_(2) omega^(2) A`
Case II: Topmost position of A

Let `T_(2)` be the tension in the elongated spring As B is at rest.
`T_(2) + R_(min) = m_(2) g`
Acceleration of A is `omega^(2) A` towards mean position (down).
`T_(2) - m_(1)g = m_(1) omega^(2) A`
Combining, we get `R_(min) = (m_(1) + m_(2)) g - m_(1) omega^(2) A`