Question

20. A bob of mass 0.1 kg hung from the ceiling of a room by a string 2 m long is set into
oscillation. The speed of the bob at its mean position is 1 ms? What is the trajectory o
the bob if the string is cut when the bob is (a) at one of its extreme positions, (b) at its
mean position? we​

Answer 1

(a) At the extreme position, the bob comes to rest so that the speed of the bob is zero.

So if the string is cut there, the bob does not advance forward but it falls downward due to gravity.

Hence the trajectory is a straight vertical line. The equation of the trajectory may be $\underline{\underline{\sf{x=0}}}$ as the horizontal displacement $\sf{x}$ is zero.

(b) At the mean position, the bob has velocity directed horizontally.

The speed is given, $\sf{u=1\ m\,s^{-1}.}$ The bob has no horizontal acceleration.

By second equation of motion, let the horizontal displacement be given by,

$\sf{\longrightarrow x=ut}$

$\sf{\longrightarrow x=t\quad\quad[\because u=1] }$

The bob has no vertical velocity at mean position, and it has vertical acceleration $\sf{g.}$

By second equation of motion, let the vertical displacement be,

$\sf{\longrightarrow y=\dfrac{1}{2}gt^2}$

$\sf{\longrightarrow\underline{\underline{y=\dfrac{1}{2}gx^2}}\quad\quad[\because x=t]}$

This is the trajectory if the string is cut at mean position.