A mass of 230 g, hanging on a spring, vertically oscillates with a period of 1 sec (the spring itself has no mass). After adding a mass, m, to the 230 g, we find that the period of oscillation of this mass-spring system becomes 2 sec. The value of m is equal to.

The added mass m= 0.7kg

Explanation:

This problem bothers on the simple harmonic motion of a spiral spring

We know that the period of a simple harmonic motion of a spring is given as

T=2π√m/k

We need to solve first for the spring constant k

Given data

Mass m =230g - - - - - kg

=230/1000= 0.230kg

Period T = 1sec

Substituting we have

1= 2*3.142√0.230/k

1=6.284√0.230/k

1/6.284=√0.230/k

Square both sides

(0.159)²=0.230/k

0.025=0.230/k

k=0.230/0.025

k= 9.2N/m

Now we find that the period of oscillation is 2 after adding mass m to 230g.. Let's solve for the new mass

Using the formula for the period T=2π√m/k

2=2*3.142√m/9.2

2=6.284√m/9.2

2/6.284=√m/9.2

Square both sides

(0.318)²=m/9.2

0.10=m/9.2

m= 0.930kg

Therefore the added mass is

0.930kg-0.230kg

The added mass m= 0.7kg

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