The figure shows a mass M attached to a series arrangement of two springs of spring constants `k_(1) and k_(2)` where `k_(2) = 2k_(1)`. If the mass M oscillates in simple harmonic motion with amplitude A, the amplitude of the point O is `(alphaA)/(beta)`. Find `beta-alpha`.

Answer

1

Explanation

Let F be the force by which mass M is stretched to the right and let A be the distance moved by it. When the mass is released, it executes S.H.M. of amplitude A. Let the extensions produced in the springs be `x_(1) and x_(2)` in springs of spring constant `k_(1) and k_(2)`, respectively.
So, `F = k_(1)x_(1) = k_(2)x_(2)`
`rArr x_(1) = (F)/(k_(1)) and x_(2) = (F)/(k_(2))`
`rArr A = x_(1) + x_(2) = F ((1)/(k_(1)) + (1)/(k_(2))) = F(k_(1) + K_(2))/(k_(1)k_(2))`
`therefore F = (k_(1)k_(2)A)/((k_(1) + k_(2)))`
The amplitude of point O = amplitude of oscillation of spring `k_(1)` which is `(k_(2) = 2k_(1))`
`x_(1) = (F)/(k_(1)) = (k_(2)A)/((k_(1) + k_(2))) = (2k_(1))/(3k_(1))A = (2)/(3) = (alpha)/(beta) A `
`therefore beta =3`
`alpha = 2`
`rArr beta -alpha = 1`