When we hang on the spring a weight mass 3 kg( 6.6 lbs), its length is 83.9 cm (33.03 in). And with weight of mass 9 kg(19.8 lbs), springs length is 142.7 cm(56.18 in). What is the springs constant? At the same spring flickers weight of 5 kg(11.2 lbs.) ?

Question

What is the springs constant?
The same spring flickers with the weight of 5 kg (11.2 lbs). What is the period of the oscillation?
What is the acceleration of the weights when the elongation is equal to 10 cm (3.93 in), 0 in or -7.87in?
At what point has the weight acceleration of 3.60 kph?

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Answer 1 · 2018-01-27T08:50:53

Suppose,the original length of the spring is #x m#

So,on applying #3 kg# as its length became #83.9 cm# or #0.839 m#

we can say, #3g = K(0.839-x)# (using #F=Kx# where, #K# is the spring constant)

Similarly for the 2nd case, #9g=K(1.427-x)#

Solving both,we get, #x= 0.545 m# and #K= 102.04 N/m#

So,on giving #5kg# weight it will extend by #(5g)/102.04# m or #0.5 m#

When elongation is #10 cm or, 0.1 m# tension in the spring is #(102.04*0.1) N# or #10.20 N#

So,net force acting downwards is #(30-10.20) N # for #3Kg#, #(90-10.20)N# for #9 Kg# and #(50-10.2) N# for #5 Kg# block.

So,their respective accelerations are, #6.6,8.86 & 7.96 m/s^2#

Suppose, at a length of r m the #5 Kg# mass has an acceleration of #a#

So, we can write, #5g - F = 5a# or, #F = 5g-5a#

Now, #F= K(r-0.545)#

So, #r-0.545 = (5(g-a))/K#

Or, #r= (5(g-a))/K +0.545#

As,we have repeatedly discussed about #F=Kx# ,so from this equation we can say #a = (Kx)/m# and for S.H.M #a= -omega ^2*x# so, #omega= sqrt(K/m)#

Hence, period of oscillation is #(2pi)/omega# i.e #2pi sqrt(m/k)#