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.

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Answers

  • Brainly User
2016-01-06T12:22:59+05:30
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
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