Question

A body of mass 10 kg is suspended by a massless coil spring of natural length 40 cm and force constant `2.0xx10^(3)Nm^(-1)`. What is stretched length of the spring.? If the body is pulled down further stretching the spring to a length 48cm and then released, what is the frequency of oscillations of the suspended mass ? `g=10ms^(2)`
A. 0.1 m
B. 0.15
C. 0.2 m
D. 0.25 m

Answer 1

Correct Answer - D
Extension `x=(mg)/(K)`
`T=2pisqrt((m)/(k)),1=2pisqrt((m)/(k))`
`pi=(1)/(4pi^(2))=(1)/(4g)`
`therefore x=(1)/(4g)xxg=(1)/(4)m`