Question

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.) ?

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?

Answer 1

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)`