. Find the magnitude of the force between two charges, where the first charge is -2µc and the second charge is 3.5µc separated b
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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
k = 244.7 N/m
Now new mass is m = 28 gram = 0.028 kg
So time period with new mass will be