Question

A vertical spring stretches 4.0 cm when a 12-g object is hung from it. The object is replaced with a block of mass 28 g that oscillates up and down in simple harmonic motion. Calculate the period of motion.

Answer 1

Time period of the osculation will be 0.0671 sec

Step-by-step explanation:

It is given a vertical spring is stretched by 4 cm

So change in length of the spring x = 4 cm = 0.04 m

Mass which is hung from it m = 12 gram = 0.012 kg

Sprig force will be equal to weight of the mass

So
`kx=mg`

`k* 0.04=0.012* 9.8`

k = 244.7 N/m

Now new mass is m = 28 gram = 0.028 kg

So time period with new mass will be

`T=2\pi \sqrt{(m)/(k)}`

`=2* 3.14 \sqrt{(0.028)/(244.7)}=0.0671sec`