Question

When a mass is suspended from a spring the latter extends over a distance of 10cm. What will be the period of oscillations of the same system if it is placed horizontal on a frictionless surface​

Answer 1

0.64 s

Step-by-step explanation:

It's period of oscillation (T) can be determined by,

T = 2
`\pi`
`\sqrt{(l)/(g) }`

Where l is the length (extension on the spring), and g the acceleration due to gravity.

But,

l = 10 cm = 0.1 m

g = 9.8 m/
`s^(2)`

Thus,

T = 2 x
`(22)/(7)`
`\sqrt{(0.1)/(9.8) }`

= 0.6350

T = 0.64 s

The period of oscillation would be 0.64 s.