Name: Properties of Solid
Uploaded: 2019-07-19
Duration: 3 min 59 s
Description: Do you know which are the important mechanical properties of solids? In this video, we will discuss the solid state of matter and its different properties.

Question

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 $P .$ What shall be the nature of the motion after the stone has reached its lowest point?

Accepted Answer

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 S.H.M. Let the stone at any instant is at y(from point $P$ ). Then

$L$

Let, $\frac{k}{m} (y - L) - g = z . . . . . . . (2)$

Differentiating equation $(2)$ with respect to $t$ , we get

$\frac{k}{m} \frac{d y}{d t} = \frac{d z}{d t} . . . . . . . . . (3)$

Again differentiating equation $(3)$ with respect to t, we get

$\frac{k}{m} \frac{d^{2} y}{d t^{2}} = \frac{d^{2} z}{d t^{2}} . . . . . . . . . (4)$

From equation $(1)$ & $(4)$ , we get

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

$\frac{m}{k} \frac{d^{2} z}{d t^{2}} + z = 0$ , From equation $(2)$

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

It is the differential equation of the second order which represents S.H.M.

$∴ \frac{d^{2} z}{d t^{2}} + ω^{2} z = 0$

where $ω$ is angular frequency so $ω = \sqrt{\frac{K}{m}}$

Solution of the above differential equation is of type

$z = A \cos (ω t + θ)$

Where $ω = \sqrt{\frac{K}{m}}$ and $θ$ is phase difference.

Substituting the value of $z$ in terms of $y$ from equation $(2)$ we get,

$\frac{k}{m} (y - L) - g = A \cos (ω t + θ) \frac{k}{m} (y - L) = A \cos (ω t + θ) + g \frac{k y}{m} = A \cos (ω t + θ) + g + \frac{k L}{m} y = \frac{m}{k} A \cos (ω t + θ) + \frac{m}{k} g + L y = (\frac{m}{k} g + L) + \frac{m}{k} A \cos (ω t + θ) y = (\frac{m}{k} g + L) + A' \cos (ω t + θ), [A' = \frac{m}{k} A]$

Thus, the stone perform S.H.M with angular frequency $ω$ about the point

$(\frac{m}{k} g + L)$ .

Important Questions on Mechanical Properties of Solids

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