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Two light strings, each of length l are ...

Two light strings, each of length `l` are fixed at points `A` and`B` on a fixed horizontal and `xy` A small are making angle `45^(@)` with the bob if the bob is displaced normal to the plane of the string and released then period of the resulting small oscillation will be

A

`2pisqrt((2sqrt(2 l))/(g))`

B

`2pi sqrt((sqrt(2l))/(g))`

C

`2pi sqrt((l)/(g))`

D

`2pi sqrt((l)/(sqrt(2g)))`

Text Solution

Verified by Experts

The correct Answer is:
D
Resulting torque on the bob `= "mg" (l)/(sqrt(2)) sin theta`
Ml of bob about axis `xy = (ml^(2))/(2)`
For small angle `theta = "mg" (l)/(sqrt(2)) sin theta`
`alpha = (tau)/(I) = (sqrt(2) g)/(l) theta`

`omega=sqrt((sqrt(2)g)/(k))rArr T=2pi sqrt((l)/(sqrt(2)g))`.
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Knowledge Check

  • Time period of small oscillation (in a verical plane normal to the plane of strings) of the bob in the arrangement shown will be

    A
    `2pisqrt((l)/(g))`
    B
    `2pisqrt((l)/(2g))`
    C
    `2pisqrt(sqrt(2l)/(g))`
    D
    `2pisqrt((2l)/(g))`

  • A simple pendulum of length 1 m the bob performs circular motion in horizontal plane if its string making an angle 60 ^(@) with the verticle , then the period of rotation of the bob will be (g=10 m//s^(2)

    A
    `2 s`
    B
    `1.4 s`
    C
    `1,98 s`
    D
    `4 s`

  • A small ball B of mass m is suspended with light inelastic string of length L from a block A of same mass in which can move on smooth horizontal surface as shown in the figure. The ball is displaced by angle theta from equilibrium position and then released. The displacement of centre of mass of A + B system till the string becomes vertical is

    A
    zero
    B
    `L(1-costheta)`
    C
    `L/2(1-costheta)`
    D
    none of these