A wire is bent an an angle `theta`. A rod of mass m can slide along the bended wire without friction as shown in Fig. A soap film is maintained in the frame kept in a vertical position and the rod is in equilibrium as shown in the figure. If rod is displaced slightly in vertical direction, then the time period of small oscillation of the rod is

Options

A. `2pisqrt((l)/(g))`
B. `2pisqrt((lcostheta)/(g))`
C. `2pisqrt((l)/(gcostheta))`
D. `2pisqrt((l)/("g"tantheta))`

Answer

A

Explanation

Let S be the surface tension of the soap film. For equilibrium of rod
`2(l+y)tan((theta)/(2))`
`mg=(F_(surface))_(1)`
`mg=(2"l"tan((theta)/(2))Sxx2`, `mg=4S"l"tan((theta)/(2))`
If the rod is displaced from its mean position by small displacement y, then restoring fornce on the rod is
`F_(rest)=-[(F_(surface))_(2)-mg]-(F_(surface))_1`
`=-[4S(l+y)tan((theta)/(2))-mg]-[4Stan((theta)/(2))y]`
`a=-(4Stan((theta)/(2)))/(m)y=(-4Stan((theta)/(2)))/((4Sl tan((theta)/(2))))`
`(d^2y)/(dt^2)=-((g)/(l))y` `because` `T=2pisqrt((l)/(g))`