Ncert Exercises & Solutions

Q.29 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?

Hint: The length of the string changes during the motion of the stone.

Consider the diagram the stone is dropped from point p.

Step 1: Find the change in length of the string.

(a) Till the stone drops through a length L, it will be in free fall. After that, the elasticity of the string will force it to do an SHM. Let the stone come to rest instantaneously at y. The loss in KE of the stone is the PE stored in the stretched string.

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

\Rightarrow m g y = \frac{1}{2} k y^{2} - k y L + \frac{1}{2} k L^{2} \Rightarrow \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}

Retain the positive sign.

∴

y = \frac{(k L + m g) + \sqrt{2 m g k L + m^{2} g^{2}}}{k}

Step 2: The maximum velocity is attained by the stone when the net force acting on it is equal to zero.

(b) In SHM, the maximum velocity is attained when the body passes through the "equilibrium position" i.e. when the instantaneous acceleration is zero. That is mg - kx = 0, where x is the extension from L.

\Rightarrow

mg = kx

Let the velocity be v. Then,

\begin{array}{l} \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} \end{array}

(from conservation of energy)
Now,

m g = k x \Rightarrow x = \frac{m g}{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}} = m g L + \frac{m^{2} g^{2}}{k} - \frac{1}{2} \frac{m^{2} g^{2}}{k}

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

∴

v^{2} = 2 g L + m g^{2} / k

v = {(2 g L + m g^{2} / k)}^{1 / 2}

Step 3: Find the equation of motion of the stone.

(c) When stone is at the lowest position i.e. at instantaneous distance y from P, then the equation of motion of the stone is:

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

∴

\frac{d^{2} z}{d t^{2}} + \frac{k}{m} z = 0

It is a differential equation of second order which represents SHM.

Comparing it with the equation,

\frac{d^{2} z}{d t^{2}} + ω^{2} z = 0

Angular frequency of harmonic motion,

ω = \sqrt{\frac{k}{m}}

The solution of the above equation will be of the type

z = A \cos (ω t + ϕ);

where

ω = \sqrt{\frac{k}{m}}

y = (L + \frac{m g}{k}) + A' \cos (ω t + ϕ)

Thus, the stone will perform SHM with an angular frequency

ω = \sqrt{k / m}

about a point

y_{0} = L + \frac{m g}{k}

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