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Mechanical Properties of Solids physics ncert solutions class 11th Physics Class 11th Physics NCERT Exemplar Solutions Class 11th Chapter Nine

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?

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Payal Gupta | Contributor-Level 10

4 months ago

This is a long answer type question as classified in NCERT Exemplar

Till the stone drops through a length L it will be in free fall. After that the elasticity of the spring will force it to a SHM. Let the stone come to rest instantaneously at y.

The loss in PE of the stone is the PE stored in the stretched string .

Mgy=1/2 k(y-L)²

Mgy = $\frac{1}{2} k y^{2} - k y L + \frac{1}{2} k L^{2}$

= $\frac{1}{2} k y^{2} - (k L + m g) y + \frac{1}{2} K L^{2} = 0$

Y= $\frac{(K L - m g) \pm \sqrt[]{{(k L + m g)}^{2} - K^{2} L^{2}}}{k} = \frac{(k L + m g) \pm \sqrt[]{2 m g k L + m^{2} g^{2}}}{k}$

b)in SHM the maximum velocity is attained when the body passes through the equilibrium position i.e when instantaneous acceleration is zero. That is mg-kx=0

so mg=kx

from the conservation of energy

$\frac{1}{2} m v^{2} + \frac{1}{2} k x^{2} = m g (L + x)$

$\frac{1}{2} m v^{2} = m g (L + x) - \frac{1}{2} k x^{2}$

mg=kx

x=mg/k

$\frac{1}{2} m v^{2} = m g (L + \frac{m g}{K}) - \frac{1}{2} k \frac{m^{2} g^{2}}{k^{2}}$

$\frac{1}{2} m v^{2} = m g L + \frac{1}{2} \frac{m^{2} g^{2}}{k}$

v²=2gL+mg²/K

v= (2gL+mg²/K)^1/2

c)when stone is at lowest position i.e at instantaneo

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This is a long answer type question as classified in NCERT Exemplar

Till the stone drops through a length L it will be in free fall. After that the elasticity of the spring will force it to a SHM. Let the stone come to rest instantaneously at y.

The loss in PE of the stone is the PE stored in the stretched string .

Mgy=1/2 k(y-L)²

Mgy = $\frac{1}{2} k y^{2} - k y L + \frac{1}{2} k L^{2}$

= $\frac{1}{2} k y^{2} - (k L + m g) y + \frac{1}{2} K L^{2} = 0$

Y= $\frac{(K L - m g) \pm \sqrt[]{{(k L + m g)}^{2} - K^{2} L^{2}}}{k} = \frac{(k L + m g) \pm \sqrt[]{2 m g k L + m^{2} g^{2}}}{k}$

b)in SHM the maximum velocity is attained when the body passes through the equilibrium position i.e when instantaneous acceleration is zero. That is mg-kx=0

so mg=kx

from the conservation of energy

$\frac{1}{2} m v^{2} + \frac{1}{2} k x^{2} = m g (L + x)$

$\frac{1}{2} m v^{2} = m g (L + x) - \frac{1}{2} k x^{2}$

mg=kx

x=mg/k

$\frac{1}{2} m v^{2} = m g (L + \frac{m g}{K}) - \frac{1}{2} k \frac{m^{2} g^{2}}{k^{2}}$

$\frac{1}{2} m v^{2} = m g L + \frac{1}{2} \frac{m^{2} g^{2}}{k}$

v²=2gL+mg²/K

v= (2gL+mg²/K)^1/2

c)when stone is at lowest position i.e at instantaneous distance Y from p, then equation of motion of the stone is

$\frac{m d^{2} y}{d t^{2}} = m g - k (y - L)$

$\frac{d^{2} y}{d t^{2}} + \frac{k}{m} (y - L) - g = 0$

Make a transformation of variables z= $\frac{k}{m} (y - L) - g$

It is differential equation of second order which represents SHM

Comparing with equation $\frac{d^{2} z}{d t^{2}} + w^{2} z = 0$

Anglular frequency of harmonic motion w = $\sqrt[]{\frac{k}{m}}$

The solution of above equation will be of the type z = Acos (wt+ $\emptyset$ ) where w = $\sqrt[]{\frac{k}{m}}$

Y= (L+mg/k)+A’cos(wt+ $\emptyset$ )

Thus , the stone will perform SHM with angular frequency w= $\sqrt[]{\frac{k}{m}}$ about a point

Y_o=L+mg/k

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