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  • A stone of mass m is tied to an elastic string of negligible mass and spring constant k. The unstretched length of the string is L and has negligible mass.

A stone of mass m is tied to an elastic string of negligible mass and spring constant k. The unstretched length of the string is L and has negligible mass. The other end of the string is fixed to a nail at a point P. Initially, the stone is at the same level as the point P. The stone is dropped vertically from point P.

a) find the distance y from the top when the mass comes to rest for an instant, for the first time

b) what is the maximum velocity attained by the stone in this drop?

c) what shall be the nature of the motion after the stone has reached its lowest point?

Answers (1)

The string is of length L, and stone is tied at P. The nail O is the point where the string is fixed. L is the height through which the stone is lifted. The stone tries to follow a path PP’ when it falls under the effect of gravity. But, due to elastic property of the string, it will follow a circular path from P to Q which gives rise to centrifugal force. The potential energy of the stone is converted to mechanical energy in the string with a spring constant K. PE of stone = Mechanical energy of the string

a) mgy = \frac{1}{2}K (y-L)^{2}

mgy = \frac{1}{2}K (y^{2}+L^{2}-2yL)

2mgy = Ky^{2} - 2KyL + KL^{2}

Ky^{2} - 2(KL+mg) y + KL^{2} = 0

 

Y=\frac{[KL+mg ]}{K}\pm \frac{\sqrt{mg(mg+2KL)}}{K}

 

b) At the lowest point, the acceleration is zero.

So, F = 0

Here the force of the string is balances by the force of gravity which makes, mg = Kx

If we assume v to be equal to maximum velocity then by the law of conservation of energy,

Kinetic energy of stone + Potential energy gained by the string = potential energy lost from P to Q’

\frac{1}{2}mv2 + \frac{1}{2} Kx^{2} = mg (L+x)

mv^{2} + Kx^{2} = 2mg(L+x)

x = \frac{mg}{K}

mv^{2} = 2mgL + \frac{2m^{2}g^{2}}{K}-\frac{m^{2}g^{2}}{K}

v = [2gL + \frac{mg^{2}}{K}]^{\frac{1}{2}}

 

c) From part a, at the lowest point

m \frac{d^{2}z}{dt^{2}} = mg - K(y-L)

\frac{d^{2}z}{dt^{2}} - g +\frac{K}{m(y-L)}= 0

z=\left [ \left ( y-L \right ) -\frac{mg}{K}\right ] {through transformation of variables}

\frac{d^{2}z}{dt^{2}} + \frac{k}{m} (z) = 0

This is a second order differential equation of the kind of simple harmonic motion.

w is the angular frequency, w=\sqrt{\frac{k}{m}}

The solution of such differential equation is of type z = A \cos (wt + \theta )

z= \left ( L+\frac{K}{m}g \right )+A^{'}\cos\left ( wt+\theta \right )

Hence, the stone undergoes simple harmonic motion about point y=0,

Hence, z_{0} = \left (L + \frac{mg}{K} \right )

Posted by

infoexpert24

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