For a two body oscillator system, prove the relation,
`T = 2pi sqrt((mu)/(k))`
where, `mu = (m_(1)m_(2))/(m_(1) + (m_(2))) =` reduced mass.

Answer

A

Explanation

A system of two bodies connected by a spring so that both are free to oscillate simple harmonically along the length of the spring constitutes a two body harmonic oscillator.

Suppose, two masses `m_(1)` and `m_(2)` are connected by a horizontal massless spring of force constant `k`, so as to be free to oscillate along the length of the spring on a frictionless horizontal surface.
Let `l_(0)` be the natural length of the spring and let `x_(1)` and `x_(2)` be the coordinates of the two masses at any instant of time. Then,
Extension of the spring `x = (x_(1) - x_(2)) - l_(0)`...(i)
For `x gt 0`, the spring force `F = kx` acts on the two masses in the direction shown in above figure.
Thus, we can write
`m_(1)(d^(2)x_(1))/(dt^(2)) = - kx`...(ii)
`m_(2)(d^(2)x_(2))/(dt_(2) = kx` ...(iii)
Multiplying Eq. (ii) by `m_(2)` and Eq. (iii) by `m_(1)` and substracting the latter from the former, we have
`m_(1) m_(2)(d^(2)x_(1))/(dt^(2)) - m_(1)m_(2)(d^(2)x_(2))/(dt^(2)) = - (m_(2) + m_(1))kx`
or `m_(1)m_(2)(d^(2)(x_(1) - x_(2)))/(dt^(2)) = - kx(m_(1) + m_(2))`
or `((m_(1)m_(2))/(m_(1) + m_(2))) (d^(2))/(dt^(2)) (x_(1) - x_(2)) = - kx` ...(iv)
Differentiating Eq. (i), twice with respect to time, we have
`(d^(2)x)/(dt^(2)) = (d^(2))/(dt^(2)) (x_(1) - x_(2))`
Also, `(m_(1)m_(2))/(m_(1) + m_(2)) = mu =` reduced mass of the two blocks
Substituting these values is Eq. (iv), we have
`mu(d^(2)x)/(dt^(2)) = - kx`
or `mua_(r) = - kx`
(Here, `a_(r) = (d^(2)x)/(dt^(2)) = (d^(2)x_(1))/(dt^(2)) - (d^(2)x_(2))/(dt^(2))` = Relative acceleration)
This, is the standard differential equation of SHM. Time period of which is
`T = 2pi sqrt|(x)/(a_(r))|` or `T = 2pi sqrt((mu)/(k))`.