Q: A bob of mass M is hung using a string of length *l*. A mass m moving with a velocity u pierces through the bob and emerges out with velocity u/3 horizontally . The frequency of small oscillations of the bob, considering A as amplitude is,

(a) $\displaystyle \frac{1}{2\pi} \sqrt{\frac{3 m u}{2 M A}} $

(b) $\displaystyle \frac{1}{2\pi} \sqrt{\frac{2 m }{3 M A}} $

(c) $ \displaystyle \frac{1}{2\pi} (\frac{2 m u}{3 M A}) $

(d) $ \displaystyle \frac{1}{2\pi} (\frac{3 m u}{2 M A}) $

Ans: (c)

Sol: mu + 0 = MV + m(u/3)

MV = 2mu/3 ⇒ V= 2mu/3M

Vmax = ωA ⇒ ω = Vmax/A = 2mu/3MA

T = 2π/ω = 2π(3MA/2mu) ⇒ ν = 1/T