Question

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

Answer 1

Answer:

The elongation of the spring is 1 cm

Explanation:

According to the problem the the length of the spring is 0.5 m . The mass of the particle connected is 0.5 kg

Now the particle is moving with the angular velocity of 2 rad /s

now let the initial length of the spring is l1 and after stretching it is l2

Now with respect to one end the spring is doing the circular motion

Then let the radius be r and r = l1+l2

Now the force applied on spring is Kx

Now the value of x is elongation on the spring

Therefore in this case,

Kl2 = mω^2(l1+l2)

Therefore l2= mω^2 x l1/ K-  mω^2

Now putting the values ,

l2= mω^2 x l1/ K-  mω^2

   = 0.5×2^2×0.5/ 100−0.5×4 =1/98m =1cm

Answer 2

Answer:

Explanation:

Here, k=100N/m,l0=0.5m,m=0.5kg

ω=2rad/s

Let elongation of the spring be l

F=kl which provides the necessary centripetal

force mω2r=mω2(l+l0)

∴Kl=mω2(l+l0)

l=mω2l0k−mω2=0.5×22×0.5100−0.5×4=198m=1cm