A body of mass m is attached to one end of a massless spring which is suspended vertically from a fixed point. The mass is held in hand so that the spring is neigther stretched nor compressed. Suddenly the support of the hand is removed. The lowest position attained by the mass during oscillation is 4cm below the point, where it was held in hand.
(a) What is the amplitude of oscillation ?
(b) Find the frequency of oscillation?

Explanation

(a) The mass m will be oscillating about a position inbetween the lowest position and the position , where it was held in hand. Therefore, the lowest position and position of hand will be extremen positions for the oscillations of mass m. The half of this distance will be amplitude of oscillation.
Thus amplitude of oscillation, `r=(4cm)/(2)=2cm`
(b) Refer to art , the frequency of oscillation of mass m suspended from a spring is `v=(1)/(2pi)sqrt((g)/(l))` where l is the extension in spring due to mass m in equilibrium position.
Here, `l=2cm, g=980 cm//s^(2)` `:. v=(1)/(2pi)sqrt((980)/(2))=(7xx7)/(2xx22)sqrt(10)=3.52s`