One end of a massless spring of spring constant 100 N/m is fixed at the centre of a frictionless horizontal table and the other end is connected to a body of mass 7 kg lying on the table at rest. The spring has a natural length of 2 m. If the spring remains horizontal and mass is made to rotate at an angular speed of 1 rad/s, then find the elongation in the spring, assuming the elongation to be small compared to the natural length of the spring.
The correct option is A 14 cm
When the body is rotating in a horizontal circular path, then the spring undergoes elongation to provide the required centripetal force for the body.
Applying the equation of circular motion along radial direction
Now let us assume the extension in spring as x metres.
∴ Spring force Fspring=kx
Substituting m= 7 kg, ω= 1 rad/s
r= natural length of spring =2 m in equation (2)
∴ x=mrω2k=7×2×(1)2100=0.14 m= 14 cm