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Find the time period T

2 votes

One rope of a swing is fixed above the other rope by b. The distance between the poles
of the swing is c. The lench l$_1$ and l$_2$ of the ropes are such that I$_1$$^2$ + l$_2$$^2$ = c$^2 $+ b$^2$ . We have to prove the period T of small oscillations of the swing as 2$\pi$ $\sqrt \frac {l_1l_2}{cg}$ . Neglect the height of swinging person in comparison with the other lengths.

I am not getting any start

asked Jan 1, 2017 in Physics Problems by koolman (4,286 points)
edited Jan 1, 2017 by koolman
Please include what you've attempted so far.
I am not getting any start . This doesn't mean  I can give my attempt to every question. As I don't know about it .
The condition $l_1+l_2=b^2+c^2$ is dimensionally incorrect. Are you sure you have copied this correctly?
Sorry , I have edited it

1 Answer

1 vote
Best answer

I have redrawn the diagram below. The mass M rotates about a point D on the fixed line AC. where angle ADM is a right angle. The component of gravity along DM is $g\cos\theta$ where $\theta$ is the angle CAB. So the swinging mass M is like a simple pendulum of of length DM in a gravitational field of strength $g\cos\theta$.

The value of $\cos\theta=c/h$ is easy to work out from $b,c$. We need to use geometry to work out the length DM.

The significance of the condition $l_1^2+l_2^2=b^2+c^2$ is that the triangle ABC is right angled, and shares the hypotenuse $AC=h$ with triangle AMC, therefore triangle AMC is also right angled and similar to ABC.

Triangle MDC is similar to triangle AMC, therefore $DM/MC=AM/AC$ hence $DM=L_1L_2/h$.

The above equations give you sufficient information to find the period of small oscillations.

answered Jan 1, 2017 by sammy gerbil (28,466 points)
selected Jan 2, 2017 by koolman
How do you get the angle made by DM by vertical as $\theta$
$AC$ is inclined at $\theta$ to the horizontal. $DM$ is perpendicular to $AC$. Therefore $DM$ is inclined at $\theta$ to the vertical.