Question

A large mass $M$ hangs stationary at the end of a light string that passes through a smooth fixed tube to a small mass $m$ that moves around in a horizontal circular path. If $\ell$ is the length of the string from $m$ to the top end of the tube and $\theta$ is angle between this part and vertical part of the string as shown in the figure, then time taken by $m$ to complete one circle is equal to:

• A. $2\pi \sqrt{\frac{\ell }{gsin\theta }}$
• B. $2\pi \sqrt{\frac{\ell }{gcos\theta }}$
• C. $2\pi \sqrt{\frac{m\ell }{gMsin\theta }}$
• D. $2\pi \sqrt{\frac{\ell m}{gM}}$

Answer 1

The correct option is D $2\pi \sqrt{\frac{\ell m}{gM}}$

Given:
Large Mass $M$ hangs stationery at the end of the light string .
FIxed tube of mass $m$ .
$l$ = length of the string
Let $T$ be the time taken to complete one circle.
x(t) = $Acos\left(wt\right)$
Maximum displacement
x(t + T ) = $Acos\left(w\right(t+T\left)\right)$
here $cos\theta =1$
hence
T = $\frac{2\times \pi }{\omega }$
$\omega =\sqrt{\left(\frac{gM}{lm}\right)}$