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

Answer:

0.64 s

Explanation:

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

T = 2$\pi$$\sqrt{\frac{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 $\frac{22}{7}$ $\sqrt{\frac{0.1}{9.8} }$

  = 0.6350

T = 0.64 s

The period of oscillation would be 0.64 s.