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

Answer:

Time period of the osculation will be 0.0671 sec  

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 $kx=mg$

$k\times 0.04=0.012\times 9.8$

k = 244.7 N/m

Now new mass is m = 28 gram = 0.028 kg

So time period with new mass will be

$T=2\pi \sqrt{\frac{m}{k}}$

$=2\times 3.14 \sqrt{\frac{0.028}{244.7}}=0.0671sec$