- Home
- Standard 11
- Physics
A stone 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 ?
Solution
Consider the diagram, the stone is dropped from point $P$.
$(a)$ Stone is in free fall upto length L. After that elasticity of string exerted force for $SHM$. Suppose, stone is at rest at instantaneous distance ' $\mathrm{y}$ '.
Loss in potential energy of stone $=$ Gain in elastic potential energy in string.
$m g y=\frac{1}{2} k(y-\mathrm{L})^{2}$
$\therefore m g y=\frac{1}{2} k y^{2}-k y \mathrm{~L}+\frac{1}{2} k \mathrm{~L}^{2}$
$\Rightarrow \frac{1}{2} k y^{2}-(k \mathrm{~L}+m g) y+\frac{1}{2} k \mathrm{~L}^{2}=0$
$y=\frac{(k \mathrm{~L}+m g) \pm \sqrt{(k \mathrm{~L}+m g)^{2}-k^{2} \mathrm{~L}^{2}}}{k}$
$\therefore y=\frac{(k \mathrm{~L}+m g) \pm \sqrt{2 m g k \mathrm{~L}+m^{2} g^{2}}}{k}$
$\therefore y=\frac{(k \mathrm{~L}+m g)+\sqrt{2 m g k \mathrm{~L}+m^{2} g^{2}}}{k}$
