One end of a massless spring of constant 100 N/m and natural length 0.5 m 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 angular velocity of 2 rad/s, then elongation of spring is

0.1 m

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10 cm

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1 cm

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0.01 cm

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Solution

The correct option is C 1 cm
Centrifugal force is balanced by spring force, thus,
$k x = m ω^{2} (l + x) \Rightarrow x = \frac{m ω^{2} l}{k - m ω^{2}} = 0.01 m = 1 c m$

Suggest Corrections

One end of a massless spring of spring constant 100 N/M and natural length 0.5 m 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.

One end of a massless spring of spring constant 100 N/m and natural length 0.5 m 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 table is made to rotate at an angular velocity of 2 rad/s, find the elongation of the spring.

Q. One end of a massless spring of spring constant