Home
Class 12
PHYSICS
One end of a massless spring of spring c...

One end of a massless spring of spring constant 100 N/m and natural length 0.5m is fixed and the other end is connected to a particle of mass 0.5 kg lying on a frictionless horizontal table. The spring remains horizontal. If the mass is made to rotate at an angular velocity of 2 rad/s, find the elongation of the spring.

Text Solution

Verified by Experts

The restoring force in the spring provides the centripetal force to the mass.
If .x. is the extension in the spring, then
`kx = m(L + x)omega^(2)`
Putting the values, we get `x=(0.5 xx 4 xx 0.5)/(100-0.5 xx 4)`
x =1 cm
Promotional Banner

Similar Questions

Explore conceptually related problems

A spring connects two particles of masses m_1 and m_2 A horizontal force F acts on m_1 Ignoring friction, when the elongation of the spring is x then:

One end of a light spring of spring constnat k is fixed to a wall and the other end is tied to a block placed on a smooth horizontal surface. In a displacement, the work done by the spring is (1)/(2)kx^(2) . The possible cases are a) the spring was initially compressed by a distance x and was finally in its natural length b) it was initially streched by a distance x and finally was in its natural length c) it was initially in its natural length and finally in a compressed position d) it was initially in its natural length and finally in a stretched position.