A body of mass m is attached ot one end of a massless spring which is suspended vertically form a fixed point. The mass is held in hand, so that the spring is neither stretched nor compressed. The lowest position attained by the mass during oscillation is 4 cm below the point, where it was held in hand.
(a) What is the amplitude of oscillation?
(b) Find the frequency of oscillation?

Explanation

(a) when the support of the hand is removed, the body oscillates about a mean position.

Suppose x is the maximum extension in the spring when it reaches the lowest point in oscillation.
Loss in PE of the block =mgx
where, m=mass of the block .....(i)
Gain in elastic potential energy of the spring
`=(1)/(2)kx^(2)" ".....(ii)`
As the two are equal, conservving the mechanical energy.
we get, `mgx=(1)/(2)kx^(2)orx=(2mg)/(g)" "....(iii)`
Now, the mean position of oscillation will be, when the block is balanced by the spring.

If x' is the extension in that case, then
`F=+kx'`
But `F=mg`
`rArrmg=+kx'`
or `x'=(mg)/(k)" ".....(iv)`
Dividing Eq. (iii) Eq. (iv),
`(x)/(x')=(2mg)/(k)/(mg)/(k)=2`
`rArrx=2x'`
But given x=4 cm (maximum extension from the unstretched position)
`:.2x'=4`
`:.x'=(4)/(2)=2cm`
But the displacement of mass from the mean position to the position when spring attains its naural length is equal to amplitude of the oscillation.
`:.A=x'=2cm`
where, A = amplitude of the motion.
(b) Time period of the oscillating system depends on mass spirng constant given by
`T=2pisqrt((m)/(k))`
It does not depend on the amplitude.
But from Eq. (iii),
`(2mg)/(k)=x" " ("maximum extension")`
`(2gm)/(k)=4cm=4xx10^(-2)cm`
`:.(m)/(k)=(4xx10^(-2))/(2g)=(2xx10^(-2))/(g)`
`:.(m)/(k)=(g)/(2xx10^(-2))`
and `v="frequency"=(1)/(T)=(1)/(2pi)sqrt((k)/(m))`
`
:.v=(1)/(2xx3.14)sqrt((4.9)/(10^(-2)))=(1)/(628)xxsqrt(4.9xx100)`
`=(10)/(628)xx221=3.51Hz.`
Note The block during oscillation connot go above the position it was released, as it is given that there is no velocity in the upward direction either by the system or external agent.