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When a mass of 40 g is attached to a vertically hanging spring it extends by 0.4 cm. Find : (See lesson 13) (i) Force constant of the spring. (ii) The

extension when 100 g weight is attached to it. (iii) The time period of oscillation of 100 g weight on it. (iv) The time period and force constant if the spring is cut in three equal parts and 100 g weight is made to oscillate on one part.

Ans 1 ) force constant of the spring -: F=mg mass = 40 gram F= (40/1000 )* 10 f=0.4 N F= kx k=spring constant x= elongation in the spring = 0.4 *10^-2 0.4= k *0.4 ^10-2 k=1/10^-2 = 100 Ans 2 the extention when 100 gram weight is attached to it then -: f= (mg) f=(100)*10/1000 f=1000/1000= 1 N f=kx k= 100 we already find the value of k only we need to find x then 1 =100 x x=10^-2 ans

Ans 3) we need to calculate its time period then use the formula and the values we find time period of the ossilation is = T=2π√m/k M = 100 GRAM K = 100 t=2π√0.1/100 ANS Ans 4 ) now time period is divided by three and k divided by 3 means T/3 = √0.1*3/100