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)`

Explanation

Here, `m=10kg, k=2.0xx10^(3)Nm^(-1)`
Natural length `l=40cm`
Extension produced in the spring dur to 10kg mass
`x=(mg)/(k)=(10xx10)/(2.0xx10^(3))=5xx10^(-2)m=5cm`
Stretched length of spriing `=l+x=40+5`
`=45cm`
When the loaded spring is stretched further, there will be no change in the frequency of oscillation of the loaded spring. If the loaded spring after further stretching is left free, it will execute linear SHM. Its frequency of oscillation
`v=(1)/(2pi)sqrt((k)/(m))=(1)/(2xx(22//7))sqrt((2.0xx10^(3))/(10))=(7xx10sqrt(2))/(2xx22)`
`=2.25Hz`