Question

A block with a mass of 3.00 kg is suspended from an ideal spring having negligible mass and stretches the spring by 0.2 m. (a) What is the force constant of the spring? (b) What is the period of oscillation of the block if it is pulled down and released ?

Answer 1

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Answer 2

Given :

The mass of a block = m = 3.00 kg

The spring stretch by = x = 0.2 meters

A block of mass m is suspended from an ideal spring having negligible mass and is stretches .

To Find :

(a) The Force constant of the spring

(b)  The period of oscillation of the block if it is pulled down and released

Solution :

From figure

( a )

At equilibrium position

    k x = m g

∵ The value of g = 9.8  m/s²

So, 0.2 m × k = 3.00 kg × 9.8  m/s²

Or,   0.2 k = 29.4  kg m/s²

∴            k = $\dfrac{29.4}{0.2}$   kg m/s²

i.e           k = 147   N/m

So, The value of spring constant = k = 147  N/m

( b )

∵             Time period of oscillation = T = 2π $\sqrt{\dfrac{l}{g} }$

And , ∵      k x = m g

Or,             k l = m g

i.e               $\dfrac{l}{g}$  = $\dfrac{m}{k}$

So,  Time period of oscillation = T = 2π $\sqrt{\dfrac{m}{k} }$

Or,                                                 T = 2 × 3.14 × $\sqrt{\dfrac{3}{147} }$

Or,                                                 T = 6.28 × $\dfrac{1}{7}$

∴   Time period of oscillation = T = 0.897  sec

Hence,

( a ) The value of spring constant is 147  N/m

( b ) The Time period of oscillation is 0.897 sec        Answer