A simple pendulum consists of a bob of mass m and a light string of length l as shown in the fig. another identical ball moving with the small velocity `v_0` collides with the pendulum's bob and sticks to it. For this new pendulum of mass 2 m, mark out the correct statements (s).

Options

A. Time period of the pendulum is `2pisqrt((l)/(g))`.
B. The equation of motion for this pendulum is `theta=(v_0)/(2sqrt(gl))sin[sqrt((g)/(l))t]`
C. The equation of motion for this pendulum is `theta=(v_0)/(2sqrt(gl))cos[sqrt((g)/(l))t]`.
D. Time period of the pendulum is `2pisqrt((2l)/(g))`.

Answer

A, B

Explanation

The time period of simple harmonic pedulum is independent of mass, so it would be same as that `T=2pisqrt((1)/(g))`. After collision, the combined mass acquires a velocity of `(v_0)/(2)`, as a result of this velocity, the mass (2m) moves up and at an angel `theta_0` (say) with vertical , it stops, this is the extreme position of bob. From work`-`energy theorem, `triangleK=W_(tat ol)`
`0-(2m)/(2)((v_0)/(2))^2=-2mgl(1-costheta_0)`
`(v_0^2)/(8gl)=1-costheta=2sin^2((theta_0)/(2))`
`sin((theta_0)/(2))=(v_0)/(4sqrt(gl))`
If `theta_0` is small `sin((theta_0)/(2))cong((theta_0)/(2))impliestheta_0=(v_0)/(2sqrt(gl))`
So, the equation of simple harmonic motion is `theta=theta_0sin(omegat)`